Chapter 17, Problem 82GQ

For each of the following cases, decide whether the pH is less than 7, equal to 7, or greater than 7.

(a) Equal volumes of 0.10 M acetic acid, CH₃CO₂H, and 0.10 M KOH are mixed.

(b) 25 mL of 0.015 M NH₃ is mixed with 12 mL of 0.015 M HCl.

(c) 150 mL of 0.20 M HNO₃, is mixed with 75 mL of 0.40 M NaOH.

(d) 25 mL of 0.45 M H₂SO₄ is mixed with 25 mL of 0.90 M NaOH.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ has to be calculated when equal volume of ${\text{CH}}_3\text{COOH}\left(0.10\text{mol}\cdot {\text{L}}^{-1}\right)$ is titrated with $\text{NaOH}\left(0.10\text{mol}\cdot {\text{L}}^{-1}\right)$.

Concept introduction:

For weak acid-strong base titration the $\text{pH}$ value can be calculated at various points before and after equivalence point. The equilibrium established during the titration of ${\text{CH}}_3\text{COOH}$ with $\text{NaOH}$. The equilibrium can be represented as,

${\text{CH}}_3\text{COOH}\left(\text{aq}\right)\text{+NaOH}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{CH}}_3\text{COONa}\left(\text{aq}\right)$

Explanation of Solution

The $\text{pH}$ calculation at equivalence point is given below.

Given:

Refer to table 16.2 in the textbook for the value of $K_{\text{a}}$.

The value of $K_{\text{a}}$ for acetic acid is $1.8\times {10}^{-5}$.

The value of $K_{\text{w}}$ for water is $1.0\times {10}^{-14}$.

The $\text{p}{K}_{\text{a}}$ value is calculated as follows;

$\text{p}{K}_{\text{a}}=-\log (K_{\text{a}})$

Substitute, $1.8\times {10}^{-5}$ for $K_{\text{a}}$.

$\begin{array}{c}\text{p}{K}_{\text{a}}=-\log (1.8\times {10}^{-5})\\ =4.74\end{array}$

Therefore, $\text{p}{K}_{\text{a}}$ values is $4.74$.

The initial concentration of ${\text{CH}}_3\text{COOH}$ is $0.100\text{mol}\cdot {\text{L}}^{-1}$.

The initial concentration of $\text{NaOH}$ is $0.100\text{mol}\cdot {\text{L}}^{-1}$.

The volume of $\text{NaOH}$ added is equal to the volume of ${\text{CH}}_3\text{COOH}$.

Therefore, let the volume of ${\text{CH}}_{\text{3}}\text{COOH}$ is $100\text{mL}$.

Thus volume of $\text{NaOH}$ added  will be $100\text{mL}$.

$\text{total}\text{volume}=\text{volume}\text{of}{\text{CH}}_3\text{COOH}\left(\text{L}\right)\text{ + volume of NaOH}\left(\text{L}\right)$

$\begin{array}{c}\text{total}\text{volume = }0.1\left(\text{L}\right)+0.1\left(\text{L}\right)\\ =0.2\text{L}\end{array}$

Therefore, total volume after reaction is $0.2\text{L}$.

The calculation of moles is done by using the expression,

$\text{Number}\text{of moles}=\text{concentration}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\cdot \text{volume}\left(\text{L}\right)$

The ICE table (1) for the reaction between $\text{NaOH}$ and ${\text{CH}}_{\text{3}}\text{COOH}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{CH}}_3\text{COOH}\left(\text{aq}\right)& +& \text{NaOH}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{CH}}_3\text{COONa}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.0100& & 0.0100& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.0100& & -0.0100& & & & 0.0100\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.0100\end{array}$

From ICE table (1),

Calculate the concentration of acetate ion after reaction.

Substitute, $0.0100\text{mol}$ for $\text{Number}\text{of moles}$ and $0.20\text{L}$ for volume in equation (4).

$\begin{array}{c}\text{concentration = }\frac{0.0100}{\text{0}\text{.20}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\\ =0.05\text{mol}\cdot {\text{L}}^{-1}\end{array}$

The concentration of acetate ion after reaction is $0.05\text{mol}\cdot {\text{L}}^{-1}$.

The acetate ion produced will undergo hydrolysis in presence of water and the reaction equilibrium is written as,

${\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{OH}}^{-}\left(\text{aq}\right)+{\text{CH}}_3\text{COOH}\left(\text{aq}\right)$

The hydrolysis equilibrium is represented in ICE table (2).

$\begin{array}{cccccccc}\text{Equation}& {\text{CH}}_3{\text{COO}}^{-}\left(\text{aq}\right)& +& {\text{H}}_2\text{O}\left(\text{l}\right)& \rightleftharpoons & {\text{OH}}^{-}\left(\text{aq}\right)& +& {\text{CH}}_3\text{COOH}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0500& & & & 0& & 0\\ \text{Change}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& -x& & & & +x& & +x\\ \text{After}\text{reaction}\left(\text{mol}\cdot {\text{L}}^{-1}\right)& 0.0500-x& & & & +x& & +x\end{array}$

From ICE table (2),

There is an approximation, that the value of $x$ is very small as comparison to $0.0500$ thus it can be neglected with respect to it.

Therefore, Concentration of acetate ion left after reaction is $0.0500\text{mol}\cdot {\text{L}}^{-1}$.

Calculate the concentration of ${\text{OH}}^{-}$ by using the equation (3).

$K_w=\left(K_{\text{a}}\right)\left(K_{\text{b}}\right)$

Rearrange it for $K_{\text{b}}$

$K_{\text{b}}=\frac{K_w}{K_{\text{a}}}$ (1)

The expression of $K_{\text{b}}$ for acetate ion from the ICE table (3) will be written as,

$K_{\text{b}}=\frac{{\left[{\text{CH}}_3\text{COOH}\right]}_{\left(\text{eq}\right)}{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_{\left(\text{eq}\right)}}$ (2)

Equate equation (5) and (6),

$\frac{K_w}{K_{\text{a}}}=\frac{{\left[{\text{CH}}_3\text{COOH}\right]}_{\left(\text{eq}\right)}{\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}}{{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_{\left(\text{eq}\right)}}$

Substitute, $1.8\times {10}^{-5}$ for $K_{\text{a}}$, $1.0\times {10}^{-14}$ for $K_{\text{w}}$, $x$ for ${\left[{\text{OH}}^{-}\right]}_{\left(\text{eq}\right)}$, $x$ for ${\left[{\text{CH}}_3\text{COOH}\right]}_{\left(\text{eq}\right)}$, $0.0500$ for ${\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_{\left(\text{eq}\right)}$.

$\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}=\frac{\left(x\right)\left(x\right)}{\left(0.0500\right)}$

$\left(0.55\times {10}^{-9}\right)=\frac{x^2}{\left(0.0500\right)}$

Rearrange for $x^2$,

$\begin{array}{c}{x}^2=\left(0.555\times {10}^{-9}\right)\left(0.0500\right)\\ =0.027\times {10}^{-9}\end{array}$

Rearrange for $x$,

$\begin{array}{c}x=\sqrt{0.27\times {10}^{-10}}\\ =0.519\times 10^{-5}\end{array}$

The value of ${\text{OH}}^{-}$ concentration is $0.519\times 10^{-5}\text{mol}\cdot {\text{L}}^{-1}$.

Calculate the value of $\text{pOH}$ by using the expression, $\text{pOH = }-\log \left[{\text{OH}}^{-}\right]$.

Substitute, $0.519\times {10}^{-5}$ for $\left[{\text{OH}}^{-}\right]$

$\begin{array}{c}\text{pOH = }-\log \left(0.519\times {10}^{-5}\right)\\ =-\left(-5.28\right)\\ =5.28\end{array}$

Therefore, the value of $\text{pOH}$ is $5.28$.

Thus, the value of $\text{pH}$ is calculated by using expression,

$\text{pH + pOH =}\text{14}$

Rearrange for $\text{pH}$,

$\text{pH =}\text{14}-\text{ pOH}$

Substitute, $5.28$ for $\text{pOH}$

$\begin{array}{c}\text{pH =}\text{14}-\text{5}\text{.28}\\ \text{= }8.72\end{array}$

Therefore, the value of $\text{pH}$ at equivalence point is $8.72$.

The value of $\text{pH}$ is $8.72$. This value is above the neutral $\text{pH}$ value $7$, thus at this point the solution will be basic in nature.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ when $25\text{mL}$,${\text{NH}}_3\left(0.015\text{mol}\cdot {\text{L}}^{-1}\right)$ is titrated with $12\text{ mL}$,$\text{HCl}\left(0.015\text{mol}\cdot {\text{L}}^{-1}\right)$ has to be calculated.

Concept introduction:

For weak base-strong acid titration the $\text{pH}$ value can be calculated at various points before and after equivalence point.

The equilibrium established during the titration of ${\text{NH}}_3$ with $\text{HCl}$. The equilibrium can be represented as,

${\text{NH}}_3\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{NH}}_4{}^{+}\left(\text{aq}\right)$

Explanation of Solution

The $\text{pH}$ calculation just before the equivalence point is done by using Henderson-Hasselbalch equation.

As the addition of $\text{HCl}$ is done there will be formation of buffer solution ${\text{NH}}_3$/${\text{NH}}_4{}^{+}$. The $\text{pOH}$ calculation for buffer solution is done by using Henderson-Hasselbalch equation.

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{conjugate}\text{acid}\right]}{\left[\text{base}\right]}$ (3)

The $\text{pH}$ value will be calculated by using expression, $\text{pH + pOH = 14}$.

Given:

Refer to table 16.2 in the textbook for the value of $K_{\text{b}}$.

The value of $K_{\text{b}}$ for ammonia is $1.8\times {10}^{-5}$.

The $\text{p}{K}_{\text{b}}$ value is calculated as follows;

$\text{p}{K}_{\text{b}}=-\log (K_{\text{b}})$

Substitute, $1.8\times {10}^{-5}$ for $K_{\text{b}}$.

$\begin{array}{c}\text{p}{K}_{\text{b}}=-\log (1.8\times {10}^{-5})\\ =4.74\end{array}$

Therefore, $\text{p}{K}_{\text{b}}$ value is $4.74$.

The initial concentration of ${\text{NH}}_3$ is $0.015\text{mol}\cdot {\text{L}}^{-1}$.

The initial concentration of $\text{HCl}$ is $0.015\text{mol}\cdot {\text{L}}^{-1}$.

The volume of ${\text{NH}}_3$ is $25\text{ mL}$.

The volume of $\text{HCl}$ is $12\text{ mL}$.

The total volume after the reaction is calculated as,

$\text{total}\text{volume}=\text{volume}\text{of}{\text{NH}}_3\left(\text{L}\right)\text{ + volume of HCl}\left(\text{L}\right)$

$\begin{array}{c}\text{total}\text{volume = }0.025\left(\text{L}\right)+0.012\left(\text{L}\right)\\ =0.037\text{L}\end{array}$

Therefore, total volume after reaction is $0.037\text{ L}$.

The calculation of moles is done by using the expression,

$\text{Number}\text{of moles}=\text{concentration}\left(\text{mol}\cdot {\text{L}}^{-1}\right)\cdot \text{volume}\left(\text{L}\right)$

The ICE table (2) for the reaction between ${\text{NH}}_3$ and $\text{HCl}$ is given below,

$\begin{array}{cccccccc}\text{Equation}& {\text{NH}}_3\left(\text{aq}\right)& +& {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)& \rightleftharpoons & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{NH}}_4{}^{+}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.000375& & 0.00018& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.00018& & -0.00018& & & & +0.00018\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0.000195& & 0& & & & 0.00018\end{array}$

From ICE table (2),

Number of moles of ammonia left after reaction are $0.000195$.

Number of moles of ${\text{H}}_3{\text{O}}^{+}$ left after reaction are $0\text{mol}$.

Number of moles of ammonium ion $\left({\text{NH}}_4{}^{+}\right)$  produced after the reaction are $0.00018$.

Concentration calculations is done by using the expression,

$\text{concentration = }\frac{\text{Number}\text{of moles}}{\text{total volume}}\left(\text{mol}\cdot {\text{L}}^{-1}\right)$

Therefore, concentration of ammonium ion $\left({\text{NH}}_4{}^{+}\right)$ after reaction is $0.0048\text{mol}\cdot {\text{L}}^{-1}$.

Concentration of ammonia after reaction is $0.0052\text{mol}\cdot {\text{L}}^{-1}$.

Calculate the $\text{pOH}$ value using equation (3).

$\text{pOH}=\text{p}{K}_{\text{b}}+\log \frac{\left[\text{conjugate}\text{acid}\right]}{\left[\text{base}\right]}$

Substitute, $0.0048$ for $\left[\text{conjugate}\text{acid}\right]$, $0.0052$ for $\left[\text{base}\right]$ and $4.74$ for $\text{p}{K}_{\text{b}}$.

$\begin{array}{c}\text{pOH}=4.74+\log \frac{\left(0.0048\right)}{\left(0.0052\right)}\\ =4.74+\log \left(0.923\right)\\ =4.74-0.0347\\ =4.70\end{array}$

Therefore, the value of $\text{pOH}$ is $4.70$.

Thus $\text{pH}$ value is calculated by using expression, $\text{pH + pOH = 14}$

Substitute, $4.70$ for $\text{pOH}$.

$\text{pH +4}\text{.70 = 14}$

Rearrange for $\text{pH}$,

$\begin{array}{c}\text{pH = 14}-\text{4}\text{.70}\\ \text{= }9.30\end{array}$

Therefore, the value of $\text{pOH}$ when $25\text{ mL}$,${\text{NH}}_{\text{3}}\left(0.015\text{mol}\cdot {\text{L}}^{-1}\right)$ is titrated with $12\text{ mL}$,$\text{HCl}\left(0.015\text{mol}\cdot {\text{L}}^{-1}\right)$ is $9.30$.

The value of $\text{pH}$ is $9.30$. This value is above the neutral $\text{pH}$ value $7$, thus at this point the solution will be basic in nature.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ when $150\text{mL}$,${\text{HNO}}_3\left(0.20\text{mol}\cdot {\text{L}}^{-1}\right)$ is titrated with $75\text{ mL}$,$\text{NaOH}\left(0.40\text{mol}\cdot {\text{L}}^{-1}\right)$ has to be calculated.

Concept introduction:

Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, ${\text{HNO}}_3$ is titrated against $\text{NaOH}$. The reaction between ${\text{HNO}}_3$ and $\text{NaOH}$ can be written as follows.

${\text{HNO}}_3\left(\text{aq}\right)+\text{NaOH}\left(\text{aq}\right)\to {\text{NaNO}}_{\text{3}}\left(\text{aq}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)$

Explanation of Solution

The $\text{pH}$ calculation for the strong acid – strong base titration is given below.

Given:

The initial concentration of ${\text{HNO}}_3$ is $0.20\text{mol}\cdot {\text{L}}^{-1}$.

The volume of ${\text{HNO}}_3$ is $150\text{ mL}$ or $0.150\text{ L}$.

The initial concentration of $\text{NaOH}$ is $0.40\text{mol}\cdot {\text{L}}^{-1}$.

The volume of $\text{NaOH}$ is $75\text{ mL}$ or $0.075\text{ L}$.

The table gives the moles of reactant and product for the reaction between ${\text{HNO}}_3$ and $\text{NaOH}$.

$\begin{array}{cccccccc}\text{Equation}& {\text{HNO}}_3\left(\text{aq}\right)& +& \text{NaOH}\left(\text{aq}\right)& \to & {\text{H}}_2\text{O}\left(\text{l}\right)& +& {\text{NaNO}}_{\text{3}}\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 0.03& & 0.03& & & & 0\\ \text{Change}\left(\text{mol}\right)& -0.03& & -0.03& & & & +0.03\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 0.03\end{array}$

After the reaction there is no moles of acid and base are present in the reaction mixture. That means all acid and base has been neutralized and titration has reached the equivalence point.

For the strong acid and strong base titration the $\text{pH}$ value is $7.0$ at the equivalence point as it is a neutral solution having water and ${\text{NaNO}}_{\text{3}}$.

 The value of $\text{pH}$ is $7.0$. Thus the solution will be neutral in nature.

Interpretation Introduction

Interpretation:

The value of $\text{pH}$ when $25\text{mL}$,${\text{H}}_2{\text{SO}}_4\left(0.45\text{mol}\cdot {\text{L}}^{-1}\right)$ is titrated with $25\text{ mL}$,$\text{NaOH}\left(0.90\text{mol}\cdot {\text{L}}^{-1}\right)$ has to be calculated.

Concept introduction:

Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, ${\text{H}}_2{\text{SO}}_4$ is titrated against $\text{NaOH}$. The reaction between ${\text{H}}_2{\text{SO}}_4$ and $\text{NaOH}$ can be written as follows.

${\text{H}}_2{\text{SO}}_4\left(\text{aq}\right)+2\text{NaOH}\left(\text{aq}\right)\to {\text{Na}}_2{\text{SO}}_4\left(\text{aq}\right)+2{\text{H}}_2\text{O}\left(\text{l}\right)$

The sulphuric acid is a dibasic acid as it has two hydrogen atoms which can be donated.

Therefore two equivalents of $\text{NaOH}$ are required to balance the reaction.

Explanation of Solution

The $\text{pH}$ calculation for the strong acid – strong base titration is given below.

Given:

The initial concentration of ${\text{H}}_2{\text{SO}}_4$ is $0.45\text{mol}\cdot {\text{L}}^{-1}$.

The volume of ${\text{H}}_2{\text{SO}}_4$ is $25\text{ mL}$ or $0.025\text{ L}$.

The initial concentration of $\text{NaOH}$ is $0.90\text{mol}\cdot {\text{L}}^{-1}$.

The volume of $\text{NaOH}$ is $25\text{ mL}$ or $0.025\text{ L}$.

The table gives the moles of reactant and product for the reaction between ${\text{H}}_2{\text{SO}}_4$ and $\text{NaOH}$.

$\begin{array}{cccccccc}\text{Equation}& {\text{H}}_2{\text{SO}}_4\left(\text{aq}\right)& +& \text{2NaOH}\left(\text{aq}\right)& \to & {\text{2H}}_2\text{O}\left(\text{l}\right)& +& {\text{Na}}_2{\text{SO}}_4\left(\text{aq}\right)\\ \text{Initial}\left(\text{mol}\right)& 22.5& & 22.5& & & & 0\\ \text{Change}\left(\text{mol}\right)& -22.5& & -22.5& & & & +22.5\\ \text{After}\text{reaction}\left(\text{mol}\right)& 0& & 0& & & & 22.5\end{array}$

After the reaction there is no moles of acid and base are present in the reaction mixture. That means all acid and base has been neutralized and titration has reached the equivalence point.

For the strong acid and strong base titration the $\text{pH}$ value is $7.0$ at the equivalence point as it is a neutral solution having water and ${\text{Na}}_2{\text{SO}}_{\text{4}}$.

The value of $\text{pH}$ is $7.0$. Thus the solution will be neutral in nature.