Chapter 16, Problem 153AP

How many grams of $\text{NaCN}$ would you need to dissolve in enough water to make exactly 250 mL of solution with a pH of 10.00?

Interpretation Introduction

Interpretation:

The amount of $\text{NaCN}$ required to make a solution of specified pH is to be calculated.

Concept introduction:

Strong electrolyte dissociates completely when added to water and weak electrolyte dissociates to a very small extent when added to water.

The reaction of the salt $\left(\text{BA}\right)$ is as follows:

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

Here, ${\text{A}}^{-}$ comes from the weak acid $\text{HA}$ and ${\text{B}}^{+}$ comes from the strong base $\text{BOH}$. The pH of this solution is determined by the $\left[{\text{OH}}^{-}\right]$.

The relationship between $K_b$, $K_a$, and ${\text{K}}_{\text{w}}$ gives the quantitative basis of the reciprocal relationship between the strength of an acid and its conjugate base or vice-versa.

$K_a\times K_b={\text{K}}_{\text{w}}$…… (1)

$K_b$ is the measure of dissociation of a base and is known as base-ionization constant that is specific at particular temperatures.

$K_b=\frac{\left[{\text{OH}}^{-}\right]\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}$…… (2)

Molarity $\left(M\right)$ of a solution is calculated as:

$M=\frac{n}{V}$…… (3)

Here, V is the volume of solution in liter and $n$ is the number of moles of solute.

pH is the measure of the acidity of a solution that depends on the concentration of hydronium ions and the temperature of the solution.

The formula to calculate the hydronium ion concentration from the pH of the solution is as follows:

$\text{pH }=-\text{ log }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\text{ , hence, }\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]=\text{ 1}0{}^{-\text{pH}}$

$\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\text{pH}}$…… (4)

Answer to Problem 153AP

Solution: $7.2\times {10}^{-3}\text{ g}$

Explanation of Solution

Given information: The volume of the solution is $250\text{ mL}$ and the pH of the solution is $10.00$.

From the given pH of the solution, calculate the concentration of ${\text{H}}_3{\text{O}}^{+}$ by equation (4) as follows:

$\begin{array}{c}\left[{\text{H}}_3{\text{O}}^{+}\right]={10}^{-\text{pH}}\\ ={10}^{-10.00}\\ =1.0\times {10}^{-10}\end{array}$

Now, calculate the concentration of ${\text{OH}}^{-}$ as follows:

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_3{\text{O}}^{+}\right]}\\ =\frac{1.0\times {10}^{-14}}{1.0\times {10}^{-10}}\\ =1.0\times {10}^{-4}\end{array}$

The salt $\text{NaCN}$ is dissociated into ${\text{Na}}^{+}$ and ${\text{CN}}^{-}$ when dissolved in water. ${\text{Na}}^{+}$ comes from a strong base $\text{NaOH}$, thus ${\text{Na}}^{+}$ does not recombine with water to form a base and remains as a free ion in the solution.

But the anion ${\text{CN}}^{-}$ comes from the weak hydrocyanic acid $\left(\text{HCN}\right)$, thus the anion recombines with water to produce the acid and hydroxide ions.

${\text{CN}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{HCN}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

From table $16.6$, the value of $K_a$ for $\text{HCN}$ is $4.9\times {10}^{-10}$.

$K_b$ of ${\text{CN}}^{-}$, which is the conjugate base of $\text{HCN}$, is calculated by equation (1) by substituting the values of $K_a$ and ${\text{K}}_{\text{w}}$ as follows:

$\begin{array}{c}\left(4.9\times {10}^{-10}\right)\times K_b=1.0\times {10}^{-14}\\ K_b=\frac{1.0\times {10}^{-14}}{4.9\times {10}^{-10}}\\ K_b=2.04\times {10}^{-5}\end{array}$

Now, prepare an equilibrium table and represent each of the species in terms of $x$ as follows:

$\begin{array}{ccccc}& {\text{CN}}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(l\right)& {\text{OH}}^{-}\left(aq\right)& \text{HCN}\left(aq\right)\\ \text{Initial concentration}\left(\text{M}\right)& x& -& 0& 0\\ \text{Change in concentration}\left(\text{M}\right)& -1.0\times {10}^{-4}& -& +1.0\times {10}^{-4}& +1.0\times {10}^{-4}\\ \text{Equilibrium concentration}\left(\text{M}\right)& x-1.0\times {10}^{-4}& -& 1.0\times {10}^{-4}& 1.0\times {10}^{-4}\end{array}$

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

$K_b=\frac{\left(1.0\times {10}^{-4}\right)\left(1.0\times {10}^{-4}\right)}{\left(x-1.0\times {10}^{-4}\right)}$

Substitute the value of $K_b$ in above equation.

$\begin{array}{c}2.04\times {10}^{-5}=\frac{\left(1.0\times {10}^{-4}\right)\left(1.0\times {10}^{-4}\right)}{\left(x-1.0\times {10}^{-4}\right)}\\ \left(x-1.0\times {10}^{-4}\right)=\frac{\left(1.0\times {10}^{-4}\right)\left(1.0\times {10}^{-4}\right)}{\left(2.04\times {10}^{-5}\right)}\\ x=\frac{\left(1.0\times {10}^{-4}\right)\left(1.0\times {10}^{-4}\right)}{\left(2.04\times {10}^{-5}\right)}+1.0\times {10}^{-4}\\ x=5.9\times {10}^{-4}\end{array}$

Thus, $\left[{\text{CN}}^{-}\right]=5.9\times {10}^{-4}\text{ M}$

The number of moles of ${\text{CN}}^{-}$ are calculatedusing equation (3) by substituting the values of molarity and volume of the solution as follows:

$\begin{array}{c}M=\frac{n}{V}\\ 5.9\times {10}^{-4}\text{ M}=\frac{n}{0.25\text{ L}}\\ n=\left(5.9\times {10}^{-4}\text{ M}\right)\left(0.25\text{ L}\right)\\ n=1.48\times {10}^{-4}\text{ mol}\end{array}$

Molar mass of $\text{NaCN}$ is $49.01\text{ g/mol}$.

Now, the mass of $\text{NaCN}$ is calculated as follows:

$\begin{array}{c}n=\frac{m}{\text{MM}}\\ m=n\left(\text{MM}\right)\\ m=\left(1.48\times {10}^{-4}\cancel{\text{mol}}\right)\left(49.01\text{ g/}\cancel{\text{mol}}\right)\\ m=7.2\times {10}^{-3}\text{ g}\end{array}$

Conclusion

The mass of $\text{NaCN}$ required to form a solution of the given pH is $7.2\times {10}^{-3}\text{ g}$.