Chapter 15, Problem 15.1KSP

The K_a for hydrocyanic acid (HCN) is 4.9 × 10 ^−l0. Determine the concentration of H⁺ in a solution that is 0.25 M in HCN.

(a) 1.2 × 10 ^−l0 M

(b) 5.0 × 10⁻⁴ M

(c) 2.2 × 10⁻⁵ M

(d) 2.5 × 10⁻⁵ M

(e) 1.1 × 10⁻⁵ M

Interpretation Introduction

Interpretation:

The concentration for ${\text{H}}^{\text{+}}$ in the dissociation of given 0.25 $\text{M}$ $\text{HCN}$ acid with given acid dissociation constant value should be determined.

Concept Introduction:

Equilibrium constant: Concentration of the products to the respective molar concentration of reactants it is called equilibrium constant. If the K value is less than one the reaction will move to the left side and the K values is higher (or) greater than one the reaction will move to the right side of reaction.

Acid dissociation constant: It is the measure which defines the strength of the acid in solution and hence it is equilibrium constant for the acid dissociation reaction represented by using ${\text{K}}_{\text{a}}$.

Answer to Problem 15.1KSP

The concentration of ${\text{H}}^{\text{+}}$ is $\left(\text{e}\right)1.1\times {10}^{-5}M$.

Explanation of Solution

Reason for correct option:

The chemical equation that represents the given conditions is as follows,

${\text{HCN}}_{\left(aq\right)}\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{\left(aq\right)}\text{+}{\text{CN}}^{\text{-}}{}_{\left(aq\right)}$

Given,

The acid dissociation constant, ${\text{K}}_{\text{a}}$ for $\text{HCN}$ is $\text{4}{\text{.9×10}}^{\text{-10}}$ and $\text{HCN}$ initial concentration is0.25 $\text{M}$.

Considering the initial concentration for products ${\text{H}}^{\text{+}}$ and ${\text{CN}}^{\text{-}}$ be 0

We set up a (ICE) equilibrium table for the given reaction as follows,

$\begin{array}{l}{\text{HCN}}_{\left(aq\right)}\stackrel{}{\to }{\text{H}}^{\text{+}}{}_{\left(aq\right)}\text{+}{\text{CN}}^{\text{-}}{}_{\left(aq\right)}\\ \text{Initial concentration: }0.2500\\ \text{After some time : }-x+x+x\\ -----------------------------\\ \text{At the end of the reaction: }0.25-xxx\end{array}$

Next using the acid dissociation formula for the given reaction we drive the value for x as follows,

$\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{H}}^{\text{+}}\right]\left[{\text{CN}}^{\text{-}}\right]}{\left[\text{HCN}\right]}\\ 4.9\times {10}^{-10}=\frac{\left[\text{x}\right]\left[\text{x}\right]}{\left[\text{0}\text{.25-x}\right]}\end{array}$

The acid dissociation constant is so small, the x in denominator of the equation gets neglected and solving further gives the value for x as follows,

$\begin{array}{l}4.9\times {10}^{-10}=\frac{\left[\text{x}\right]\left[\text{x}\right]}{\left[\text{0}\text{.25-x}\right]}\\ 4.9\times {10}^{-10}=\frac{{\text{x}}^{\text{2}}}{0.25}\\ {\text{x}}^{\text{2}}=4.9\times {10}^{-10}\times 0.25\end{array}$

$\begin{array}{l}=1.225\times {10}^{-10}\\ \text{x}\text{=1}\text{.1}\times {10}^{-5}M\end{array}$

Therefore, the above calculations clearly says that the concentration of ${\text{H}}^{\text{+}}$ ions is $\text{1}\text{.1}\times {10}^{-5}M$ since the final concentration of ${\text{H}}^{\text{+}}$ at the end of the reaction is x and the x value is equal to $\text{1}\text{.1}\times {10}^{-5}M$.

Hence, the correct option is option (e).

Reason for in-correct options:

Consider all options other than the correct one.

$\begin{array}{l}\left(\text{a}\right)\text{1}{\text{.2×10}}^{\text{-10}}\text{M}\\ \left(\text{b}\right)\text{5}{\text{.0×10}}^{\text{-4}}\text{M}\\ \left(\text{c}\right)2.{\text{2×10}}^{\text{-5}}\text{M}\\ \left(\text{d}\right)\text{2}{\text{.5×10}}^{\text{-5}}\text{M}\end{array}$

These options show us that they do not contain the correct value which is obtained from the calculation using acid dissociation constant formula and the ice table for the given chemical reaction.

Hence, the incorrect options are option (a), (b), (c) and (d).

Conclusion

The correct option that contains the value for the concentration of ${\text{H}}^{\text{+}}$ in the given acid dissociation reaction was determined by using the acid dissociation formula.