Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M NH 3 ( K b = 1.8 × 10 −5 ) with 0.100 M HCl.

Question

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Textbook Question

Chapter 14, Problem 63E

Repeat the procedure in Exercise 61, but for the titration of 25.0 mL of 0.100 M NH₃ (K_b = 1.8 × 10⁻⁵) with 0.100 M HCl.

Expert Solution & Answer

Interpretation Introduction

Interpretation: The titration of $NH3$ with different volumes of $HCl$ is given. The $pH$ of each solution is to be calculated and then graph of $pH$ versus volume of titrant added is to be plotted.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The $pH$ of each solution is to be calculated and graph of $pH$ versus volume of titrant added.

Answer to Problem 63E

Answer

The value of $pH$ of solution when $0.0 mL$ $HCl$ has been added is. $11.11_$ .

The value of $pH$ of solution when $4.0 mL$ $HCl$ has been added is. $9.85_$ .

The value of $pH$ of solution when $8.0 mL$ $HCl$ has been added is. $9.61_$ .

The value of $pH$ of solution when $12.5 mL$ $HCl$ has been added is. $9.26_$ .

The value of $pH$ of solution when $20.0 mL$ $HCl$ has been added is. $8.66_$ .

The value of $pH$ of solution when $24.0 mL$ $HCl$ has been added is. $7.86_$ .

The value of $pH$ of solution when $24.5 mL$ $HCl$ has been added is. $7.56_$ .

The value of $pH$ of solution when $24.9 mL$ $HCl$ has been added is. $6.86_$ .

The value of $pH$ of solution when $25.0 mL$ $HCl$ has been added is. $5.25_$ .

The value of $pH$ of solution when $25.1 mL$ $HCl$ has been added is $3.69_$

The value of $pH$ of solution when $26.0 mL$ $HCl$ has been added is $2.69_$

The value of $pH$ of solution when $28.0 mL$ $HCl$ has been added is $2.22_$

The value of $pH$ of solution when $30.0 mL$ $HCl$ has been added is $2.04_$

The graph between $pH$ and volume of $HCl$ added is shown in Figure 2.

Explanation of Solution

Explanation

Given

The concentration of $NH3$ is $0.100 M$

The concentration of $HCl$ is $0.100 M$ .

The volume of $NH3$ is $25.0 mL$ .

The volume of $HCl$ is $0.0 mL$ , $4.0 mL$ , $8.0 mL$ , $12.5 mL$ , $20.0 mL$ , $24.0 mL$ , $24.5 mL$ , $24.9 mL$ , $25.0 mL$ , $25.1 mL$ , $26.0 mL$ , $28.0 mL$ , $30.0 mL$

The value of $Kb$ of $NH3$ is $1.8×10−5$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $100 mL$ into $L$ is done as,

$100 mL=100×0.001 L=0.100 L$

The reaction is represented as,

$NH3+H2O⇌NH4++OH−$

Make the ICE table for the reaction between $H2NNH2$ and $H2O$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.10000Change:−x+xxFinal moles:0.100−xxx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]$

Since, value of $Kb$ is very small, hence, $(0.100−x)$ is taken as $0.100$ .

Simplify the above equation,

$1.8×10−5=(x)(x)(0.100) Mx=0.0013 M$

It is the concentration of $OH−$ .

The $pOH$ of the solution is shown below.

$pOH=−log[OH−]$ (1)

Where,

$[OH−]$ is the concentration of Hydroxide ions.

Substitute the value of $[OH−]$ in the above equation as,

$pOH=−log[OH−]=−log(0.0013)=2.89$

The value of $pOH$ of the solution is $2.89$ .

The relationship between $pOH$ is given as,

$pH+pOH=14$ (2)

Substitute the value of $pOH$ in the above equation.

$pH+pOH=14pH+2.89=14pH=11.11_$

The value of $pH$ of solution when $0.0 mL$ $HCl$ has been added is. $11.11_$ .

The volume of $HCl$ is $4.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $4.0 mL$ into $L$ is done as,

$4.0 mL=4.0×0.001 L=0.004 L$

The concentration of any species is given as,

$Concentration=Number of molesVolume of solution in litres$ (3)

Rearrange the above equation to obtain the value of number of moles.

$Number of moles=Concentration×Volume of solution in litres$ (4)

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.004 L=0.0004 moles$

Substitute the value of concentration and volume of $NH3$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.025 L=0.0025 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.00040Change in moles −0.0004−0.0004+0.0004Equilibrium moles0.00210.00040.0004$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.004 L=0.029 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0021 moles0.029 L=0.05 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0004 moles0.029 L=0.013 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.050.013Change:−x+xxFinal moles:0.05−x0.013+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.013+x)(x)(0.05−x) M$

Since, value of $Kb$ is very small, hence, $(0.05−x)$ is taken as $(0.05)$ and $(0.013+x)$ is taken as $0.013$ .

Simplify the above equation,

$1.8×10−5=(0.013+x)(x)(0.05−x) M1.8×10−5=(0.013)(x)(0.05) Mx=7.0×10−5 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the equation (1).

$pOH=−log[OH−]=−log(7.0×10−5)=4.15$

The value of $pOH$ of the solution is $4.15$ .

Substitute the value of $pOH$ in the equation (2)

$pH+pOH=14pH+4.15=14pH=9.85_$

The value of $pH$ of solution when $4.0 mL$ $HCl$ has been added is. $9.85_$ .

The volume of $HCl$ is $8.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $8.0 mL$ into $L$ is done as,

$8.0 mL=8.0×0.001 L=0.008 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.008 L=0.0008 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.00080Change in moles −0.0004−0.0008+0.0008Equilibrium moles0.00170.00080.0008$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.008 L=0.033 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0017 moles0.033 L=0.0515 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0008 moles0.033 L=0.0242 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.05150.02420Change:−x+xxFinal moles:0.0515−x0.0242+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.0242+x)(x)(0.0515−x) M$

Since, value of $Kb$ is very small, hence, $(0.0515−x)$ is taken as $(0.0515)$ and $(0.0242+x)$ is taken as $0.0242$ .

Simplify the above equation,

$1.8×10−5=(0.0242+x)(x)(0.0515−x) M1.8×10−5=(0.0242)(x)(0.0515) Mx=4.0×10−5 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the equation (1).

$pOH=−log[OH−]=−log(4.0×10−5)=4.39$

The value of $pOH$ of the solution is $4.39$ .

Substitute the value of $pOH$ in the equation (2)

$pH+pOH=14pH+4.39=14pH=9.61_$

The value of $pH$ of solution when $8.0 mL$ $HCl$ has been added is. $9.61_$ .

The volume of $HCl$ is $12.5 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $12.5 mL$ into $L$ is done as,

$12.5 mL=12.5×0.001 L=0.0125 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.0125 L=0.00125 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.001250Change in moles −0.00125−0.00125+0.00125Equilibrium moles0.001250.001250.00125$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.0125 L=0.0375 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00125 moles0.0375 L=0.033 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00125 moles0.0375 L=0.033 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.0330.0330Change:−x+xxFinal moles:0.033−x0.033+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.033+x)(x)(0.033−x) M$

Since, value of $Kb$ is very small, hence, $(0.033−x)$ is taken as $(0.033)$ and $(0.033+x)$ is taken as $0.033$ .

Simplify the above equation,

$1.8×10−5=(0.033+x)(x)(0.033−x) M1.8×10−5=(0.033)(x)(0.033) Mx=1.8×10−5 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the equation (1).

$pOH=−log[OH−]=−log(1.8×10−5)=4.74$

The value of $pOH$ of the solution is $4.74$ .

Substitute the value of $pOH$ in the above equation as,

$pH+pOH=14pH+4.74=14pH=9.26_$

The value of $pH$ of solution when $12.5 mL$ $HCl$ has been added is. $9.26_$ .

The volume of $HCl$ is $20.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $20.0 mL$ into $L$ is done as,

$20.0 mL=20.0×0.001 L=0.02 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.02 L=0.002 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.0020Change in moles −0.002−0.002+0.002Equilibrium moles0.00050.0020.002$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.02 L=0.045 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0005 moles0.045 L=0.011 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.002 moles0.045 L=0.044 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.0110.0440Change:−x+xxFinal moles:0.011−x0.044+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.044+x)(x)(0.011−x) M$

Since, value of $Kb$ is very small, hence, $(0.011−x)$ is taken as $(0.011)$ and $(0.044+x)$ is taken as $0.044$ .

Simplify the above equation,

$1.8×10−5=(0.044+x)(x)(0.011−x) M1.8×10−5=(0.044)(x)(0.011) Mx=4.5×10−6 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the equation (1).

$pOH=−log[OH−]=−log(4.5×10−6)=5.34$

The value of $pOH$ of the solution is $5.34$ .

Substitute the value of $pOH$ in the equation (2).

$pH+pOH=14pH+5.34=14pH=8.66_$

The value of $pH$ of solution when $20.0 mL$ $HCl$ has been added is. $8.66_$ .

The volume of $HCl$ is $24.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $24.0 mL$ into $L$ is done as,

$24.0 mL=24.0×0.001 L=0.024 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.024 L=0.0024 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.00240Change in moles −0.0024−0.0024+0.0024Equilibrium moles0.00010.00240.0024$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.024 L=0.049 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0001 moles0.049 L=0.002 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0024 moles0.049 L=0.05 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.0020.050Change:−x+xxFinal moles:0.002−x0.05+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.05+x)(x)(0.002−x) M$

Since, value of $Kb$ is very small, hence, $(0.002−x)$ is taken as $(0.002)$ and $(0.05+x)$ is taken as $0.05$ .

Simplify the above equation,

$1.8×10−5=(0.05+x)(x)(0.002−x) M1.8×10−5=(0.05)(x)(0.002) Mx=7.2×10−7 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the above equation as,

$pOH=−log[OH−]=−log(7.2×10−7)=6.14$

The value of $pOH$ of the solution is $6.14$ .

Substitute the value of $pOH$ in the above equation as,

$pH+pOH=14pH+6.14=14pH=7.86_$

The value of $pH$ of solution when $24.0 mL$ $HCl$ has been added is. $7.86_$ .

The volume of $HCl$ is $24.5 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $24.5 mL$ into $L$ is done as,

$24.5 mL=24.5×0.001 L=0.0245 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.0245 L=0.00245 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.002450Change in moles −0.00245−0.00245+0.00245Equilibrium moles0.000050.002450.00245$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.0245 L=0.0495 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00005 moles0.0495 L=0.001 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00245 moles0.0495 L=0.05 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.0010.050Change:−x+xxFinal moles:0.001−x0.05+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.05+x)(x)(0.001−x) M$

Since, value of $Kb$ is very small, hence, $(0.001−x)$ is taken as $(0.001)$ and $(0.05+x)$ is taken as $0.05$ .

Simplify the above equation,

$1.8×10−5=(0.05+x)(x)(0.001−x) M1.8×10−5=(0.05)(x)(0.001) Mx=3.6×10−7 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the equation (1).

$pOH=−log[OH−]=−log(3.6×10−7)=6.44$

The value of $pOH$ of the solution is $6.44$ .

Substitute the value of $pOH$ in the equation (2).

$pH+pOH=14pH+6.44=14pH=7.56_$

The value of $pH$ of solution when $24.5 mL$ $HCl$ has been added is. $7.56_$ .

The volume of $HCl$ is $24.9 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $24.9 mL$ into $L$ is done as,

$24.9 mL=24.9×0.001 L=0.0249 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.0249 L=0.00249 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.002490Change in moles −0.00249−0.00249+0.00249Equilibrium moles0.000010.002490.00249$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.0249 L=0.0499 L$

Substitute the value of number of moles of $NH3$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00001 moles0.0499 L=0.0002 M$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00249 moles0.0499 L=0.05 M$

Make the ICE table for the dissociation reaction of $NH3$ .

$NH3+H2O⇌NH4++OH−Initial moles:0.00020.050Change:−x+xxFinal moles:0.0002−x0.05+xx$

The equilibrium ratio for the given reaction is,

$Kb=[NH4+][OH−][NH3]$

Substitute the calculated concentration values in the above expression.

$Kb=[NH4+][OH−][NH3]1.8×10−5=(0.05+x)(x)(0.0002−x) M$

Since, value of $Kb$ is very small, hence, $(0.0002−x)$ is taken as $(0.0002)$ and $(0.05+x)$ is taken as $0.05$ .

Simplify the above equation,

$1.8×10−5=(0.05+x)(x)(0.0002−x) M1.8×10−5=(0.05)(x)(0.0002) Mx=7.2×10−8 M$

It is the concentration of $OH−$ .

Substitute the value of $[OH−]$ in the equation (1).

$pOH=−log[OH−]=−log(7.2×10−8)=7.14$

The value of $pOH$ of the solution is $7.14$ .

Substitute the value of $pOH$ in the equation (2).

$pH+pOH=14pH+7.14=14pH=6.86_$

The value of $pH$ of solution when $24.9 mL$ $HCl$ has been added is. $6.86_$ .

The volume of $HCl$ is $25.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $25.0 mL$ into $L$ is done as,

$25.0 mL=25.0×0.001 L=0.025 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.025 L=0.0025 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.00250Change in moles −0.0025−0.0025+0.0025Equilibrium moles0.00.00.0025$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.025 L=0.05 L$

Substitute the value of number of moles of $NH4+$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0025 moles0.05 L=0.05 M$

Make the ICE table for the reaction of $NH4+$ with water.

$NH4++H2O⇌NH3+H3O+Initial moles:0.0500Change:−x+xxFinal moles:0.05−xxx$

The relationship between $Ka$ and $Kb$ is given as,

$Ka×Kb=10−14$

Substitute the value of $Kb$ in above equation.

$Ka×Kb=10−14Ka×1.8×10−5=10−14Ka=5.6×10−10$

The equilibrium ratio for the given reaction is,

$Ka=[NH3][H3O+][NH4+]$

Substitute the calculated concentration values in the above expression.

$Ka=[NH3][H3O+][NH4+]5.6×10−10=(x)(x)(0.05−x) M$

Since, value of $Ka$ is very small, hence, $(0.05−x)$ is taken as $(0.05)$ .

Simplify the above equation,

$5.6×10−10=(x)(x)(0.05−x) M5.6×10−10=(x)(x)(0.05) Mx=5.5×10−6 M$

It is the concentration of $H3O+$ .

The $pH$ of the solution is shown below.

$pH=−log[H3O+]$ (5)

Where,

$[H3O+]$ is the concentration of Hydronium ions in the solution.

Substitute the value of $[H3O+]$ in the above equation.

$pH=−log[H3O+]=−log(5.5×10−6)=5.25_$

The value of $pH$ of solution when $25.0 mL$ $HCl$ has been added is. $5.25_$ .

The volume of $HCl$ is $25.1 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $25.1 mL$ into $L$ is done as,

$25.1 mL=25.1×0.001 L=0.0251 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.025 L=0.0025 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.002510Change in moles −0.0025−0.0025+0.0025Equilibrium moles0.00.000010.0025$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.0251 L=0.0501 L$

Substitute the value of number of moles of $HCl$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.00001 moles0.0501 L=0.0002 M$

The concentration of $H+$ is $0.0002 M$ .

Substitute the value of $[H+]$ in the equation (5).

$pH=−log[H+]=−log(0.0002)=3.69_$

The value of $pH$ of solution when $25.1 mL$ $HCl$ has been added is $3.69_$ .

The volume of $HCl$ is $26.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $26.0 mL$ into $L$ is done as,

$26.0 mL=26.0×0.001 L=0.026 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.026 L=0.0026 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.00260Change in moles −0.0025−0.0025+0.0025Equilibrium moles0.00.00010.0025$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.026 L=0.051 L$

Substitute the value of number of moles of $HCl$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0001 moles0.051 L=0.002 M$

Substitute the value of $[H+]$ in the above equation as,

$pH=−log[H+]=−log(0.002)=2.69_$

The value of $pH$ of solution when $26.0 mL$ $HCl$ has been added is $2.69_$

The volume of $HCl$ is $28.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $28.0 mL$ into $L$ is done as,

$28.0 mL=28.0×0.001 L=0.028 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.028 L=0.0028 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.00280Change in moles −0.0025−0.0025+0.0025Equilibrium moles0.00.00030.0025$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.028 L=0.053 L$

Substitute the value of number of moles of $HCl$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0003 moles0.053 L=0.006 M$

The concentration of $H+$ is $0.006 M$ .

Substitute the value of $[H+]$ in the equation (5).

$pH=−log[H+]=−log(0.006)=2.22_$

The value of $pH$ of solution when $28.0 mL$ $HCl$ has been added is $2.22_$

The volume of $HCl$ is $30.0 mL$ .

The conversion of $mL$ into $L$ is done as,

$1 mL=0.001 L$

Hence the conversion of $30.0 mL$ into $L$ is done as,

$30.0 mL=30.0×0.001 L=0.03 L$

Substitute the value of concentration and volume of $HCl$ in equation (4) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.03 L=0.003 moles$

Make the ICE table for the reaction between $NH3$ and $HCl$ .

$NH3+HCl⇌NH4ClInitial moles 0.00250.0030Change in moles −0.0025−0.0025+0.0025Equilibrium moles0.00.00050.0025$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $=Volume of NH3+Volume of HCl=0.025 L+0.03 L=0.055 L$

Substitute the value of number of moles of $HCl$ and final volume of solution in equation (1).

$Concentration=Number of molesVolume of solution in litres=0.0005 moles0.055 L=0.009 M$

The concentration of $H+$ is $0.009 M$ .

Substitute the value of $[H+]$ in the equation (5).

$pH=−log[H+]=−log(0.009)=2.04_$

The value of $pH$ of solution when $30.0 mL$ $HCl$ has been added is $2.04_$

The values of $pH$ obtained are shown in the below table.

Volume of $HCl$ in $mL$	$pH$
$0.0$	$11.11$
$4.0$	$9.85$
$8.0$	$9.61$
$12.5$	$9.26$
$20.0$	$8.66$
$24.0$	$7.86$
$24.5$	$7.56$
$24.9$	$6.86$
$25.0$	$5.25$
$25.1$	$3.69$
$26.0$	$2.69$
$28.0$	$2.22$
$30.0$	$2.04$

Figure 1

The graph between $pH$ and volume of $HCl$ added is shown below.

EBK CHEMISTRY: AN ATOMS FIRST APPROACH, Chapter 14, Problem 63E

Figure 2

Conclusion

Conclusion

The amount of species present in the solution during titration depends on the volume of titrant added in the solution and this further defines the value of $pH$ of the solution

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