The titration of Nitric acid with different volumes of NaOH is given. The pH of each solution is to be calculated and the graph between pH and volume of base 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.

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Answer 1

Question

Chapter 15, Problem 102AE

Interpretation Introduction

Interpretation: The titration of Nitric acid with different volumes of $NaOH$ is given. The $pH$ of each solution is to be calculated and the graph between $pH$ and volume of base 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.

Expert Solution & Answer

Answer to Problem 102AE

Answer

The value of $pH$ of solution when $0.0 mL$ $NaOH$ has been added is. $1.0_$ .

The value of $pH$ of solution when $4.0 mL$ $NaOH$ has been added is. $1.14_$ .

The value of $pH$ of solution when $8.0 mL$ $NaOH$ has been added is. $1.28_$ .

The value of $pH$ of solution when $12.5 mL$ $NaOH$ has been added is. $1.48_$ .

The value of $pH$ of solution when $20.0 mL$ $NaOH$ has been added is. $1.95_$ .

The value of $pH$ of solution when $24.0 mL$ $NaOH$ has been added is. $2.69_$ .

The value of $pH$ of solution when $24.5 mL$ $NaOH$ has been added is. $3.0_$ .

The value of $pH$ of solution when $24.9 mL$ $NaOH$ has been added is. $3.69_$ .

The value of $pH$ of solution when $25.0 mL$ $NaOH$ has been added is. $7.0_$ .

The value of $pH$ of solution when $25.1 mL$ $NaOH$ has been added is. $10.35_$ .

The value of $pH$ of solution when $26.0 mL$ $NaOH$ has been added is. $11.31_$ .

The value of $pH$ of solution when $28.0 mL$ $NaOH$ has been added is. $11.78_$ .

The value of $pH$ of solution when $30.0 mL$ $NaOH$ has been added is. $11.96_$ .

The graph between $pH$ and volume of base added is shown in Figure 1.

Explanation of Solution

Explanation

The value of $pH$ of solution when $0.0 mL$ $NaOH$ has been added is. $1.0_$ .

Given:

The concentration of $HNO3$ acid is $0.100 M$

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

The volume of $HNO3$ is $25.0 mL$ .

The volume of $NaOH$ is $0.0 mL$ .

When no base is added, then solution contains only the strong acid $HNO3$ . Therefore, $pH$ is given by concentration of $H+$ only.

The $pH$ of a solution is shown below.

$pH=−log[H+]$ (1)

Where,

$[H+]$ is the concentration of ions present in a solution.

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

$pH=−log[H+]=−log(0.100)=1.0_$

The value of $pH$ of solution when $0.0 mL$ $NaOH$ has been added is. $1.0_$ .

Explanation

The concentration of $H+$ is $0.0724 M$ .

Given

The volume of $NaOH$ 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,

$Concentration=Number of molesVolume of solution in litres$ (2)

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

$Number of moles=Concentration×Volume of solution in litres$ (3)

Substitute the value of concentration and volume of $HNO3$ in equation (3) as,

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

Substitute the value of concentration and volume of $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.0004Change 0.0025−0.0004−0.0004Final moles0.00210$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.004 L=0.029 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

$Concentration=Number of molesVolume of solution in litres=0.0021 moles0.029 L=0.0724 M$

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $4.0 mL$ $NaOH$ has been added is. $1.14_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.0724)=1.14_$

The value of $pH$ of solution when $4.0 mL$ $NaOH$ has been added is. $1.14_$ .

Explanation

The concentration of $H+$ is $0.0515 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.0008Change 0.0025−0.0008−0.0008Final moles0.00170$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.008 L=0.033 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

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

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $8.0 mL$ $NaOH$ has been added is. $1.28_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.0515)=1.28_$

The value of $pH$ of solution when $8.0 mL$ $NaOH$ has been added is. $1.28_$ .

Explanation

The concentration of $H+$ is $0.033 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.00125Change 0.0025−0.00125−0.00125Final moles0.001250$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.0125 L=0.0375 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

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

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $12.5 mL$ $NaOH$ has been added is. $1.48_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.033)=1.48_$

The value of $pH$ of solution when $12.5 mL$ $NaOH$ has been added is. $1.48_$ .

Explanation

The concentration of $H+$ is $0.011 M$ .

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.002Change 0.0025−0.002−0.002Final moles0.00050$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.02 L=0.045 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

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

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $20.0 mL$ $NaOH$ has been added is. $1.95_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.011)=1.95_$

The value of $pH$ of solution when $20.0 mL$ $NaOH$ has been added is. $1.95_$ .

Explanation

The concentration of $H+$ is $0.002 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.0024Change 0.0025−0.0024−0.0024Final moles0.00010$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.024 L=0.049 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

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

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $24.0 mL$ $NaOH$ has been added is. $2.69_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.002)=2.69_$

The value of $pH$ of solution when $24.0 mL$ $NaOH$ has been added is. $2.69_$ .

Explanation

The concentration of $H+$ is $0.001 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.00245Change 0.0025−0.00245−0.00245Final moles0.000050$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.0245 L=0.0495 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

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

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $24.5 mL$ $NaOH$ has been added is. $3.0_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.001)=3.0_$

The value of $pH$ of solution when $24.5 mL$ $NaOH$ has been added is. $3.0_$ .

Explanation

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

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.00249Change 0.0025−0.00249−0.00249Final moles0.000010$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.0249 L=0.0499 L$

Substitute the value of number of moles of $HNO3$ and final volume of solution in equation (2).

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

It is the concentration of $H+$ .

Explanation

The value of $pH$ of solution when $24.9 mL$ $NaOH$ has been added is. $3.69_$ .

Substitute the value of $[H+]$ in the equation (1).

$pH=−log[H+]=−log(0.0002)=3.69_$

The value of $pH$ of solution when $24.9 mL$ $NaOH$ has been added is. $3.69_$ .

Explanation

The value of $pH$ of solution when $25.0 mL$ $NaOH$ has been added is. $7.0_$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.0025Change 0.0025−0.0025−0.0025Final moles0.00$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.025 L=0.05 L$

As all the moles have been neutralized, therefore the value of $pH$ is $7.0_$ .

The value of $pH$ of solution when $25.0 mL$ $NaOH$ has been added is. $7.0_$ .

Explanation

The concentration of $OH−$ is $0.00022 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

$Number of moles=Concentration×Volume of solution in litres=0.100 M×0.0251 L=0.00251 moles$

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.00251Change 0.0025−0.00250.00251−0.0025Final moles00.00001$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.0251 L=0.0451 L$

Substitute the value of number of moles of $NaOH$ and final volume of solution in equation (2).

$Concentration=Number of molesVolume of solution in litres=0.00001 moles0.0451 L=0.00022 M$

It is the concentration of $OH−$ .

Explanation

The value of $pOH$ of solution when $25.1 mL$ $NaOH$ has been added is. $3.65$ .

The $pOH$ of the solution is shown below.

$pOH=−log[OH−]$ (4)

Where,

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

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

$pOH=−log[OH−]=−log(0.00022)=3.65$

Explanation

The value of $pH$ of solution when $25.1 mL$ $NaOH$ has been added is. $10.35_$ .

The relationship between $pOH$ is given as,

$pH+pOH=14$

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

$pH+pOH=14pH+3.65=14pH=10.35_$

The value of $pH$ of solution when $25.1 mL$ $NaOH$ has been added is. $10.35_$ .

Explanation

The concentration of $OH−$ is $0.002 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.0026Change 0.0025−0.00250.0026−0.0025Final moles00.0001$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.026 L=0.051 L$

Substitute the value of number of moles of $NaOH$ and final volume of solution in equation (2).

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

It is the concentration of $OH−$ .

Explanation

The value of $pOH$ of solution when $26.0 mL$ $NaOH$ has been added is. $2.69$ .

Substitute the value of $[OH−]$ in the equation (4).

$pOH=−log[OH−]=−log(0.002)=2.69$

The value of $pOH$ of solution when $26.0 mL$ $NaOH$ has been added is. $2.69$ .

Explanation

The value of $pH$ of solution when $26.0 mL$ $NaOH$ has been added is. $11.31_$ .

The relationship between $pOH$ is given as,

$pH+pOH=14$ (5)

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

$pH+pOH=14pH+2.69=14pH=11.31_$

The value of $pH$ of solution when $26.0 mL$ $NaOH$ has been added is. $11.31_$ .

Explanation

The concentration of $OH−$ is $0.006 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.0028Change 0.0025−0.00250.0028−0.0025Final moles00.0003$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.028 L=0.053 L$

Substitute the value of number of moles of $NaOH$ and final volume of solution in equation (2).

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

It is the concentration of $OH−$ .

Explanation

The value of $pOH$ of solution when $28.0 mL$ $NaOH$ has been added is. $2.22$ .

Substitute the value of $[OH−]$ in the equation (4).

$pOH=−log[OH−]=−log(0.006)=2.22$

The value of $pOH$ of solution when $28.0 mL$ $NaOH$ has been added is. $2.22$ .

Explanation

The value of $pH$ of solution when $25.1 mL$ $NaOH$ has been added is. $11.78_$ .

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

$pH+pOH=14pH+2.22=14pH=11.78_$

The value of $pH$ of solution when $28.0 mL$ $NaOH$ has been added is. $11.78_$ .

Explanation

The concentration of $OH−$ is $0.009 M$ .

Given

The volume of $NaOH$ 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 $NaOH$ in equation (3) as,

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

Make the table for the reaction between $HNO3$ and $NaOH$ .

$H++OH−→H2OInitial moles 0.00250.003Change 0.0025−0.00250.003−0.0025Final moles00.0005$

Total volume of solution $=Volume of HNO3+Volume of NaOH=0.025 L+0.03 L=0.055 L$

Substitute the value of number of moles of $NaOH$ and final volume of solution in equation (2).

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

It is the concentration of $OH−$ .

Explanation

The value of $pOH$ of solution when $30.0 mL$ $NaOH$ has been added is. $2.04$ .

Substitute the value of $[OH−]$ in the equation (4).

$pOH=−log[OH−]=−log(0.009)=2.04$

The value of $pOH$ of solution when $30.0 mL$ $NaOH$ has been added is. $2.04$ .

Explanation

The value of $pH$ of solution when $30.0 mL$ $NaOH$ has been added is. $11.96_$ .

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

$pH+pOH=14pH+2.04=14pH=11.96_$

The value of $pH$ of solution when $30.0 mL$ $NaOH$ has been added is. $11.96_$ .

Explanation

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

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

$pH$	Volume of $NaOH$ in $mL$
$1.0$	$0.0$
$1.14$	$4.0$
$1.28$	$8.0$
$1.48$	$12.5$
$1.95$	$20.0$
$2.69$	$24.0$
$3.0$	$24.5$
$3.69$	$24.9$
$7.0$	$25.0$
$10.35$	$25.1$
$11.31$	$26.0$
$11.78$	$28.0$
$11.96$	$30.0$

Table 1

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

WebAssign for Zumdahl/Zumdahl/DeCoste's Chemistry, 10th Edition [Instant Access], Single-Term, Chapter 15, Problem 102AE

Figure 1

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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