Chapter 8, Problem 178CP

(a)

Interpretation Introduction

Interpretation:

The pH of solution before any 0.10 M NaOH has been added needs to be determined.

Concept Introduction :

The pH of the acids mixture is determined from the stronger acid. H₂SO₄ is a strong diprotic acid. So, it completely dissociates into H+ and HSO₄- ions. Again, in the solution, HSO₄- becomes the strongest acid and it also dissociates into H+ and SO₄²- ions.

Answer to Problem 178CP

The pH of the mixture before adding 0.1 M NaOH is 1.36.

Explanation of Solution

The ICE table for the dissociation of HSO₄- ion is shown below

	HSO4-	H+	SO4-
I	0.05	0.05	0
C	-X	+X	+X
E	0.05-X	0.05+X	X

However, equilibrium constant, K_a for HSO₄- is 1.2×10^-2 and pK_a = 1.92

  $\begin{array}{l}{K}_a=\frac{[S{O}_4^{2-}\text{}][H^{+}]}{[HS{O}_4^{-}\text{}]}\text{}\\ 1.2\times {10}^{-2}=\frac{(X)(0.05+X)\text{}}{0.05-X}\\ X=0.00722\end{array}$

[H⁺] at equilibrium = 0.05 - 0.00722 = 0.04278 M

So, the pH of the mixture before adding 0.1 M NaOH

  $\begin{array}{l}pH=-log\left(0.04278\right)\\ =1.36\end{array}$

(b)

Interpretation Introduction

Interpretation:

The pH of solution after a total of 100.0 mL of 0.10 M NaOH has been added needs to be determined.

Concept Introduction :

  $Number\text{ of moles of acid = concentration}\times \text{Volume of solution}$

Answer to Problem 178CP

The pH of the solution after adding 100 mL of 0.1 M NaOH is 1.52.

Explanation of Solution

Consider 1 L of acids mixture is titrated with 100 mL of NaOH.

Now,

  $\begin{array}{l}\begin{array}{l}Thenumberofmolesofacid=0.04278\frac{mol}{\sout{L}}\times 1\sout{L}\hfill \\ =0.04278mol\hfill \\ \hfill \\ ThenumberofmolesofNaOH=0.10\frac{mol}{\sout{L}}\times 100\times {10}^{-3}\sout{L}\hfill \end{array}\\ =0.01mol\end{array}$

The OH- anions will be completely consumed by the reaction and H₃O+ ion will remain.

So, the remaining number of moles of H₃O+ ion in the solution

  $\begin{array}{l}Numberofmolesof{H}_3{O}^{+}ioninthesolution\hfill \\ =0.04278mol-0.01mol\hfill \\ =0.03278mol\hfill \end{array}$

  $\begin{array}{l}Thetotalvolumeofthesolution=1000mL+100mL=1100mL=1.1L\\ Theconcentrationof{H}_3{O}^{+}ion\text{ =}\frac{0.03278 mol\text{}}{1.1 L}=0.0298M\end{array}$

So, the pH of the solution after adding 100 mL of 0.1 M NaOH

  $\begin{array}{l}pH=-log\left(0.0298\right)\\ =\mathrm{1.52.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The pH of solution after a total of 300.0 mL of 0.10 M NaOH has been added needs to be determined.

Concept Introduction :

The pH of solution is calculated as follows:

  $pH=-log\left[H^{+}\right]$

Here, $\left[H^{+}\right]$ is hydrogen ion concentration.

Answer to Problem 178CP

The pH of the solution after adding 300 mL of 0.1 M NaOH is 2.01.

Explanation of Solution

Now, adding the additional 200 mL of NaOH in the previous solution.

The number of moles of acid after adding 100 mL of NaOH = 0.03278 mol

  $ThenumberofmolesofNaOH=0.10mol/\sout{L}\times 200\times {10}^{-3}\sout{L}=0.02mol$

The H₃O+ ions will be completely consumed by the reaction and H₃O+ ion remains.

So, the remaining number of moles of H₃O+ ion after adding 300 mL of NaOH

  $\begin{array}{l}\begin{array}{l}Theremainingnumberofmolesof{H}_3{O}^{+}ion\\ =0.03278mol-0.02mol\end{array}\hfill \\ =0.01278mol\hfill \end{array}$

The total volume of the solution = 1100 mL + 200 mL = 1300 mL = 1.3 L

  $\begin{array}{l}Theconcentrationof{H}_3{O}^{+}ion\\ =\frac{0.01278 mol}{1.3L}\text{}\\ =0.00983M\end{array}$

So, the pH of the solution after adding 300 mL of 0.1 M NaOH

  $\begin{array}{l}\begin{array}{l}pH=-log\left(0.00983\right)\\ =2.01\end{array}\hfill \\ \hfill \end{array}$

(d)

Interpretation Introduction

Interpretation:

The pH of solution after a total of 500.0 mL of 0.10 M NaOH has been added needs to be determined.

Concept Introduction :

The pH of solution is calculated as follows:

  $pH=-log\left[H^{+}\right]$

Here, $\left[H^{+}\right]$ is hydrogen ion concentration.

Also, pOH =14-( -log[H⁺])

Answer to Problem 178CP

The pH of the solution after adding 500 mL of 0.1 M NaOH is 11.683.

Explanation of Solution

Again, adding additional 200 mL of NaOH in the previous solution.

Now, the number of moles of acid after adding 300 mL of NaOH = 0.01278 mol

  $\begin{array}{l}ThenumberofmolesofNaOH\\ =0.10mol/\sout{L}\times 200\times {10}^{-3}\sout{L}\\ =0.02mol\end{array}$

Here,

  $n_{O{H}^{-}}\text{}>n_{H^{+\text{}}},$

This means that all the H3O+ ions in the solution are completely consumed and hence, the resulting solution becomes basic.

So,

  $n_{O{H}^{-}}\text{}-n_{H^{+\text{}}}=0.00722mol$

  $\begin{array}{l}Thetotalvolumeofthesolution=1300mL+200mL\\ =1500mL\\ =1.5L\end{array}$

The concentration of OH- ion

  $\begin{array}{l}Concentration==\frac{0.00722 mol}{1.5L}\text{}\\ =0.00481\text{ mol/L}\\ \text{=0}\text{.00481}M\end{array}$

So, the pH of the solution after adding 500 mL of 0.1 M NaOH

  $\begin{array}{l}pH=14-\left[-log\left(0.00481\right)\right]\hfill \\ \begin{array}{l}=14-2.317\\ =\mathrm{11.683.}\end{array}\hfill \end{array}$