Chapter 19.3, Problem 34LC

Interpretation Introduction

Interpretation: The classification of the given solution in terms of concentration and strength needs to be done.

Concept Introduction: For any acid dissociation reaction, the acid dissociation constant can be calculated as follows:

  $K_a=\frac{\left[{\text{A}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[\mathrm{HA}\right]}$

Here, the concentration of dissociated species can be calculated using the ICE table.

Explanation of Solution

The given solution is a 15 M solution of an acid with an acid dissociation constant $7.5\times {10}^{-3}$ .

The strength of the solution can be determined by calculating the degree of dissociation.

The acid dissociation reaction can be represented as follows:

  $\text{HA}+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{A}}^{-}$

Here, the concentration of acid $\left(\text{HA}\right)$ is 15 M.

The ICE table can be represented as follows:

  $\begin{array}{l}\text{ HA}+{\text{H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}+{\text{A}}^{-}\\ I\text{ C - - -}\\ C-\text{C}\alpha \text{ - C}\alpha \text{ C}\alpha \\ E\text{ C}-\text{C}\alpha \text{ - C}\alpha \text{ C}\alpha \end{array}$

Now, the expression for the acid dissociation constant will be:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$

Substitute the values,

  $K_a=\frac{\left(\text{C}\alpha \right)\left(\text{C}\alpha \right)}{\text{ C}\left(\text{1}-\alpha \right)}$

Here, the value of C is the initial concentration of HA that is 15 M, and the value of $K_a$ is $7.5\times {10}^{-3}$ .

Now, substitute all the values in the expression for the acid dissociation constant.

  $K_a=\frac{\left(\text{C}\alpha \right)\left(\text{C}\alpha \right)}{\text{ C}\left(\text{1}-\alpha \right)}$

Or,

  $\begin{array}{c}7.5\times {10}^{-3}=\frac{15{\alpha }^2}{\left(\text{1}-\alpha \right)}\\ 15{\alpha }^2=7.5\times {10}^{-3}-7.5\times {10}^{-3}\alpha \\ 15{\alpha }^2+7.5\times {10}^{-3}\alpha -7.5\times {10}^{-3}=0\end{array}$

Solve the above quadratic equation as follows:

  $\begin{array}{c}x=\frac{-7.5\times {10}^{-3}\pm \sqrt{{\left(7.5\times {10}^{-3}\right)}^2-4\left(15\right)\left(-7.5\times {10}^{-3}\right)}}{2\left(15\right)}\\ =\frac{-7.5\times {10}^{-3}\pm \sqrt{0.00005625+0.45}}{2\left(15\right)}\\ =\frac{-0.0075\pm 0.67}{30}\\ =+0.0221,-0.0226\end{array}$

The value of x cannot be negative thus, the correct value of x or $\alpha$ is $0.0221$

The given solution is a concentrated weak acid. The solution is concentrated because the molarity of the solution is high and it is a weak acid because it is not completely dissociated into hydronium ion and conjugate base.