Chapter 16, Problem 86QP

What are the concentrations of ${\text{HSO}}_{\text{4,}}^{\text{–}}{\text{ SO}}^{\text{2–}}{}_4{\text{, and H}}_{\text{3}}{\text{O}}^{\text{–}}\text{ in a 0}\text{.20 }M{\text{ KHSO}}_{\text{4}}\text{ solution?}$ (Hint: ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{) is a strong acid; }{K}_{\text{a}}{\text{for HSO}}_{\text{4}}^{\text{-}}{\text{ = 103 × 10}}^{\text{-2}}\text{.}$

Interpretation Introduction

Interpretation:

The concentrations of ${\text{HSO}}_4{}^{-},{\text{ SO}}_4{}^{2-}{\text{, and H}}_3{\text{O}}^{+}$ in the solution that contains ${\text{KHSO}}_4$ are to be determined.

Concept introduction:

Salt is a strong electrolyte that dissociates completely when added to water.

When a salt contains an anion that comes from a strong acid, the anion does not recombine with water to produce the acid. It exists as a free ion in the solution and does not affect the pH of the solution. The cation, if it comes from a strong base, does not recombine and is present in the solution as a free ion without having any effect on the pH of the solution.

For the diprotic acid, the first ionization takes place as:

${\text{H}}_2{\text{A}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{HA}}^{-}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{a1}}$ is the measure of dissociation of the first proton of an acid and is known as the first acid-ionization constant, which is specific at a particular temperature.

${\text{K}}_{\text{a1}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{HA}}^{-}\right]}{\left[{\text{H}}_2\text{A}\right]}$ …… (1)

The second ionization of the diprotic acid takes place as:

${\text{HA}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}+{\text{A}}^{-}{}_{\left(\text{aq}\right)}$

${\text{K}}_{\text{a2}}$ is the measure of dissociation of the second proton of an acid and is known as the second acid-ionization constant, which is specific at a particular temperature.

${\text{K}}_{\text{a2}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[{\text{HA}}^{-}\right]}$ …… (2)

Percent ionization is the percentage of acid that gets dissociated upon addition in water. It depends on the hydronium ion concentration.

$\%\text{ dissociation}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[\text{HA}\right]}_{\text{o}}}\times 100\%$ …… (3)

Here, ${\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}$ is the hydronium ion concentration at equilibrium and ${\left[\text{HA}\right]}_{\text{o}}$ is the original acid concentration.

Answer to Problem 86QP

Solution:

$\begin{array}{c}\left[{\text{HSO}}_{\text{4}}{}^{-}\right]=0.16M\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=0.045M\\ \left[{\text{SO}}_4{}_{2-}\right]=0.045M\end{array}$

Explanation of Solution

Given information:

The concentration of potassium hydrogen sulphate $\left({\text{KHSO}}_4\right)$ at $25{}^{\circ }\text{C}$ is $0.20M$.

The salt ${\text{KHSO}}_4$ is dissociated into ${\text{K}}^{+}$ and ${\text{HSO}}_4{}^{-}$ when dissolved in water. ${\text{K}}^{+}$ comes from a strong base $\text{KOH}$. Thus, ${\text{K}}^{+}$ does not recombine with water to form a base and remains as a free ion in the solution.

Also, the anion ${\text{HSO}}_4{}^{-}$ comes from sulfuric acid $\left({\text{H}}_2{\text{SO}}_4\right)$, which is a strong acid. Thus, the aniondo not recombine with water to produce the acid. But it contains a proton to be donated and thus, acts as a weak acid and produces hydroxide ions and its conjugate base, ${\text{SO}}_4{}^{2-}$, according to the reaction:

${\text{HSO}}_4{{}^{-}}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}\rightleftharpoons {\text{H}}_3{\text{O}}^{+}{}_{\left(\text{aq}\right)}+{\text{SO}}_4{{}^{2-}}_{\left(\text{aq}\right)}$

The $K_{\text{a}}$ for this species, $\left({\text{HSO}}_4{}^{-}\right)$ is given as $1.3\times {10}^{-2}$.

Since, there is complete dissociation of the salt to form ions,

$\begin{array}{c}\left[{\text{K}}^{+}\right]=0.20M\\ \left[{\text{HSO}}_4{}^{-}\right]=0.20M\end{array}$

Prepare an equilibrium table and represent each of the species in terms of $y$ as:

$\begin{array}{ccccc}& {\text{HSO}}_4{{}^{-}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{SO}}_4{{}^{-}}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.20& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -y& -& +y& +y\\ \text{Equilibrium concentration}\left(M\right)& 0.20-y& -& y& y\end{array}$

Now, substitute these concentrations in equation (2) as:

${\text{K}}_{\text{a}}=\frac{\left(y\right)\left(y\right)}{\left(0.20-y\right)}$

Since the value of ${\text{K}}_{\text{a}}$ is small, the amount of acid dissociated is less. Therefore, $\left(0.20-y\right)$ can be approximated as $0.20$. Now, substitute the value of ${\text{K}}_{\text{a}}$ in the above equation as:

$\begin{array}{c}1.3\times {10}^{-2}=\frac{\left(y\right)\left(y\right)}{0.20}\\ y^2=2.6\times {10}^{-3}\\ y=\sqrt{2.6\times {10}^{-3}}\\ y=5.1\times {10}^{-2}\end{array}$

Thus, ${\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}_{\left(\text{eq}\right)}=5.1\times {10}^{-2}M$, ${\left[{\text{HSO}}_{\text{4}}{}^{-}\right]}_{\left(\text{O}\right)}=0.20M$

Calculate the percent dissociation from equation (3) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{5.1\times {10}^{-2}}{\text{0}\text{.20}}\times 100\%\\ =25.495\%\end{array}$

Since, the percent dissociation is more than $5\%$, the approximation taken is not valid. Thus, solve again for the value of $y$ as:

$\begin{array}{c}\text{1}\text{.3}\times {\text{10}}^{-2}=\frac{\left(y\right)\left(y\right)}{\left(0.20-y\right)}\\ \left(2.6\times {10}^{-3}\right)-\left(\text{1}\text{.3}\times {\text{10}}^{-2}\right)y=y^2\\ 0=y^2+\left(1.3\times {10}^{-2}\right)y-\left(2.6\times {10}^{-3}\right)\\ y=-0.058,\text{ 0}\text{.045}\end{array}$

Since, concentration cannot be negative so, $y=0.045$.

Thus,

$\begin{array}{c}\left[{\text{HSO}}_{\text{4}}{}^{-}\right]=\left(0.20-0.045\right)M\approx 0.16M\\ \left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=0.045M\\ \left[{\text{SO}}_4{}_{2-}\right]=0.045M\end{array}$

Conclusion

The concentrations of ${\text{HSO}}_4{}^{-},{\text{ SO}}_4{}^{2-}{\text{, and H}}_3{\text{O}}^{+}$ in the salt solution of ${\text{KHSO}}_4$ are $0.16M,0.045M,\text{ and }0.045M$, respectively.