The value of p K b of trimethylamine is given to be 4.19 at 25 ° C . The pH of the solution is to be calculated after addition of 10.0 , 20.0 , and 30.0 ml of acid. Concept introduction: The value of pH is calculated by the use of concentration of H + and OH − ions in the solution. pH is the numeric value which defines acidity and basicity of the solution. To determine: The pH of the solution after addition of 10.0 , 20.0 , and 30.0 ml of acid.

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Answer 1

Question

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

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Answer 2

Question

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

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Answer 3

Question

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

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Answer 4

Question

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

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Answer 5

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Answer 6

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Answer 7

Chapter 16, Problem 16.116QP

Interpretation Introduction

Interpretation: The value of $pKb$ of trimethylamine is given to be $4.19$ at $25° C$ . The $pH$ of the solution is to be calculated after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Concept introduction: The value of $pH$ is calculated by the use of concentration of $H+$ and $OH−$ ions in the solution. $pH$ is the numeric value which defines acidity and basicity of the solution.

To determine: The $pH$ of the solution after addition of $10.0, 20.0, and 30.0 ml$ of acid.

Answer 8

Expert Solution & Answer

Answer to Problem 16.116QP

Answer

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Explanation

Given

The value of $pKb$ of solution is $4.19$ .

Volume of trimethylamine, $V1$ is $25.0 ml$ .

Concentration of trimethylamine, $M1$ is $0.100 M$ .

Concentration of $HCl$ , $M2$ is $0.125 M$ .

The addition of $HCl$ in a solution, $V2$ is $10 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid, BH+=0.125×10=1.25 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The value of $mmol$ of base after reaction is calculated by subtract the equation (2) and (4),

$mmol of base after reaction, B=2.5−1.25=1.25 mmol$

The value of $mmol$ of conjugate that is formed during the reaction is $1.25$ . The given reaction forms a buffer solution.

The entire base reacts with acid, so the total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Where,

$Vacid$ is the volume of acid.
$Vbase$ is the volume of base.

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of $pOH$ is calculated by the formula,

$pOH=pKb+log(BH+/B)$

Substitute the value of $pKb, BH+ and B$ in above equation,

$pOH=4.19+log(1.25/1.25)=4.19$

The value of $pH$ is calculated by the formula,

$pH=14−pOH$ (5)

Substitute the value of $pOH$ in equation (5).

$pH=14−4.19=9.81_$

Therefore, the value of $pH$ is $9.81_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $20 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×20=2.5 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

At this point value of $mmol$ of acid becomes equal to the value of $mmol$ of base.

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=20+25=45 ml$

The value of conjugate formation is equal to $2.5 mmol$ .

The value of conjugate concentration is calculated by the formula,

$[BH+]=mmol of conjugate saltVtotal=2.545=0.05555 M$

The equation of reaction of conjugate formed with water is given as,

$BH++H2O→H3O++B$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]=KwKb$ (6)

The value of $Kw$ is $10−14$ .

Substitute the value of $Kw and Kb$ in equation (6).

$Ka=10−1410−4.19=1.548×10−10$

So, $[H3O+]=x=[B][BH+]=M−x=0.05555−x$

The value of $Ka$ is calculated by the formula,

$Ka=[H3O+][B][BH+]$ (7)

Where,

$Ka$ is equilibrium constant.
$[H3O+]$ is the concentration of an acid.
$[BH+]$ is the concentration of conjugate salt.

Substitute the value of $[H3O+], [B] and [BH+]$ in equation (7).

$1.548×10−10=x×x(0.5555−x)x2=1.548×10−10×(0.5555−x)x2=1.548×10−10×0.5555−1.548×10−10x0=x2−0.859914×10−10+1.548×10−10x$

Simplify the above expression,

$x=−1.548×10−10±(−1.548×10−10)−4(1×0.859914×10−10)2x=−1.548×10−10±2.39×10−20+3.43×10−102x=−1.548×10−10+5.82×10−302x=2.93×10−6$

The value of $[H+]$ is equal to $x$ .

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(2.93×10−6)=5.53_$

Therefore, the value of $pH$ is $5.53_$ .

Similarly,

The addition of $HCl$ in a solution, $V2$ is $30 ml$ .

The equation for given reaction is,

$(CH3)3N+HCl→(CH3)3NH+Cl−$

The value of $mmol$ of acid is calculated by the formula,

$mmol of acid=M2V2$ (1)

Substitute the value of $M2 and V2$ in equation (1).

$mmol of acid=0.125×30=3.75 mmol$ (2)

The value of $mmol$ of base is calculated by the formula,

$mmol of base=M1V1$ (3)

Substitute the value of $M1 and V1$ in equation (1).

$mmol of base=0.1×25=2.5 mmol$ (4)

The total volume of solution is calculated by addition of acidic and basic substances.

$VTotal=Vacid+Vbase$ (5)

Substitute the values of $Vacid and Vbase$ in equation (5).

$VTotal=30+25=55 ml$

The acid is excess and the value of $mmol$ of acid left is calculated by the formula,

$mmol of remained acid=mmol of acid−mmol of base$ (6)

Substitute the values of $mmol$ of acid and base in equation (6).

$mmol of remained acid=3.75−2.5=1.25 mmol$

The concentration of acid is calculated by the formula,

$[acid]=mmol of acidVtotal of acid$ (7)

Substitute the value of $mmol of acid and Vtotal of acid$ in equation (7).

$[acid]=mmol of acidVtotal of acid=1.2555=0.0227 M$

The value of $pH$ is calculated by the formula,

$pH=−log[H+]$ (8)

Substitute the value of $[H+]$ in equation (8).

$pH=−log(0.227 M)=1.65$

Therefore, the value of $pH$ is $1.65_$ .

Answer 12

Conclusion

Conclusion

The value of $pH$ after addition of $10 ml$ of acid is $9.81_$ .

The value of $pH$ after addition of $20 ml$ of acid is $5.53_$ .

The value of $pH$ after addition of $30 ml$ of acid is $1.65_$ .

Answer 13

Ch. 16.1 - Prob. 1PE Ch. 16.2 - Prob. 2PE Ch. 16.2 - Prob. 3PE Ch. 16.3 - Prob. 4PE Ch. 16.3 - Prob. 5PE Ch. 16.4 - Prob. 6PE Ch. 16.6 - Prob. 7PE Ch. 16.6 - Prob. 8PE Ch. 16.7 - Prob. 9PE Ch. 16.8 - Prob. 10PE

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