Chapter 16, Problem 57QP

Determine the percent ionization of the following solutions of formic acid at $\text{25°C}:(\text{a})0.016M,(\text{b})5.7\times {10}^{-4}M,(\text{c})1.75M.$

Interpretation Introduction

Interpretation:

The percent ionization of the solution prepared by the dissolution of formic acid is to be determined.

Concept introduction:

The ionization of the weak acid takes place as:

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

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

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

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

$\%\text{ ionization}=\frac{{\left[{\text{H}}_3{\text{O}}^{+}\right]}_{\text{eq}}}{{\left[\text{HA}\right]}_{\text{o}}}\times 100\%$ …… (2)

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 57QP

Solution:

$10\%$

$42.1\%$

$0.97\text{ \%}$

Explanation of Solution

Given information:

a) $0.016M$

Refer Table $16.6$ for the acid ionization constant of formic acid as ${\text{K}}_{\text{a}}=1.7\times {10}^{-4}$.

When a weak acid is dissolved in water, only partial dissociation takes place into ${\text{H}}_3{\text{O}}^{+}$ ions and its conjugate base. Thus, the concentration of ${\text{H}}_3{\text{O}}^{+}$ is determined by ${\text{K}}_{\text{a}}$ of the acid.

The reaction of the weak acid $\left(\text{HCOOH}\right)$ is depicted as:

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

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

$\begin{array}{ccccc}& {\text{HCOOH}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{HCOO}}^{-}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 0.016& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.016-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as:

${\text{K}}_{\text{a}}=\frac{\left(x\right)\left(x\right)}{\left(0.016-x\right)}$

Since the value of ${\text{K}}_{\text{a}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(0.016-x\right)$ can be approximated as $0.016$. Now, substitute the value of ${\text{K}}_{\text{a}}$ in the above equation as:

$\begin{array}{c}1.7\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{0.016}\\ x^2=\left(1.7\times {10}^{-4}\right)0.016\\ x=\sqrt{2.72\times {10}^{-6}}\\ x=1.65\times {10}^{-3}\end{array}$

Thus, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.65\times {10}^{-3}M$

Calculate the percent dissociation from equation (2) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{1.65\times {10}^{-3}}{\text{0}\text{.016}}\times 100\%\\ =10.31\%\end{array}$

Since the percent dissociation is more than $5\%$, the approximation is not valid. Again, solve for $x$ without taking the approximation as:

$\begin{array}{c}1.7\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{\left(0.016-x\right)}\\ x^2=\left(2.72\times {10}^{-6}\right)-\left(1.7\times {10}^{-4}\right)x\\ x=\sqrt{2.72\times {10}^{-6}}\\ x=1.57\times {10}^{-3},-1.7\times {10}^{-3}\end{array}$

The value of $x$ cannot be negative. Thus, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.57\times {10}^{-3}M$.

Again, calculate the percent dissociation with this value as:

$\begin{array}{c}\%\text{ dissociation}=\frac{1.57\times {10}^{-3}}{\text{0}\text{.016}}\times 100\%\\ =10\%\end{array}$

Therefore, the percent dissociation is $10\%$.

b) $\text{5}\text{.7}\times {\text{10}}^{-4}M$

Refer Table $16.6$ for the acid ionization constant of formic acid as ${\text{K}}_{\text{a}}=1.7\times {10}^{-4}$.

When a weak acid is dissolved in water, only partial dissociation takes place into ${\text{H}}_3{\text{O}}^{+}$ ions and its conjugate base. Thus, the concentration of ${\text{H}}_3{\text{O}}^{+}$ is determined by ${\text{K}}_{\text{a}}$ of the acid.

The reaction of the weak acid $\left(\text{HCOOH}\right)$ is depicted as:

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

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

$\begin{array}{ccccc}& {\text{HCOOH}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{HCOO}}^{-}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& \text{5}\text{.7}\times {\text{10}}^{-4}& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& \left(\text{5}\text{.7}\times {\text{10}}^{-4}\right)-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as:

${\text{K}}_{\text{a}}=\frac{\left(x\right)\left(x\right)}{\left(\left(\text{5}\text{.7}\times {\text{10}}^{-4}\right)-x\right)}$

The value of ${\text{K}}_{\text{a}}$ is very small, but the given concentration is small too. Therefore, the amount of acid dissociated is taken as $\left(\left(\text{5}\text{.7}\times {\text{10}}^{-4}\right)-x\right)$ only. Substitute the value of ${\text{K}}_{\text{a}}$ in the above equation as:

$\begin{array}{c}1.7\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{\left(\left(\text{5}\text{.7}\times {\text{10}}^{-4}\right)-x\right)}\\ x^2=\left(9.69\times {10}^{-8}\right)-\left(1.7\times {10}^{-4}\right)x\\ x=2.4\times {10}^{-4},-\text{4}\text{.1}\times {10}^{-4}\end{array}$

The concentration cannot be negative. Thus, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=2.4\times {10}^{-4}M$.

Calculate the percent dissociation from equation (2) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{2.4\times {10}^{-4}}{5.7\times {10}^{-4}}\times 100\%\\ =42.1\%\end{array}$

Therefore, the percent dissociation is $42.1\%$.

c) $\text{1}\text{.75 }M$

Refer Table $16.6$ for the acid ionization constant of formic acid as ${\text{K}}_{\text{a}}=1.7\times {10}^{-4}$.

When a weak acid is dissolved in water, only partial dissociation takes place into ${\text{H}}_3{\text{O}}^{+}$ ions and its conjugate base. Thus, the concentration of ${\text{H}}_3{\text{O}}^{+}$ is determined by ${\text{K}}_{\text{a}}$ of the acid.

The reaction of the weak acid $\left(\text{HCOOH}\right)$ is depicted as:

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

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

$\begin{array}{ccccc}& {\text{HCOOH}}_{\left(\text{aq}\right)}& {\text{H}}_2{\text{O}}_{\left(\text{l}\right)}& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\left(\text{aq}\right)}& {\text{HCOO}}^{-}{}_{\left(\text{aq}\right)}\\ \text{Initial concentration}\left(M\right)& 1.75& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 1.75-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (1) as:

${\text{K}}_{\text{a}}=\frac{\left(x\right)\left(x\right)}{\left(1.75-x\right)}$

Since the value of ${\text{K}}_{\text{a}}$ is very small, the amount of acid dissociated is less. Therefore, $\left(1.75-x\right)$ can be approximated as $1.75$. Now, substitute the value of ${\text{K}}_{\text{a}}$ in the above equation as:

$\begin{array}{c}1.7\times {10}^{-4}=\frac{\left(x\right)\left(x\right)}{1.75}\\ x^2=\left(1.7\times {10}^{-4}\right)1.75\\ x=\sqrt{2.975\times {10}^{-4}}\\ x=1.7\times {10}^{-2}\end{array}$

Thus, $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=1.7\times {10}^{-2}M$

Calculate the percent dissociation from equation (2) as:

$\begin{array}{c}\%\text{ dissociation}=\frac{1.7\times {10}^{-2}}{\text{1}\text{.75}}\times 100\%\\ =0.97\%\end{array}$

Therefore, the percent dissociation is $0.97\%$.