Answer
(b) 11.28
Work Step by Step
1. $CN^-$ is the conjugate base of a weak acid; therefore, it is a weak base.
2. Find its $K_b$:
$$K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{ 4.9 \times 10^{-10} } = 2.0 \times 10^{-5} $$
3. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ CN^- ]& [ HCN ]& [ OH^- ]\\
Initial& 0.175 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.175 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
4. Write the expression for $K_b$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_b = \frac{[Products]}{[Reactants]} = \frac{[ HCN ][ H^+ ]}{[ CN^- ]}$$
$$K_b = \frac{(x)(x)}{[ CN^- ]_{initial} - x}$$
5. Assuming $ 0.175 \gt\gt x$:
$$K_b = \frac{x^2}{[ CN^- ]_{initial}}$$
$$x = \sqrt{K_b \times [ CN^- ]_{initial}} = \sqrt{ 2.0 \times 10^{-5} \times 0.175 }$$
$x = 1.9 \times 10^{-3} $
6. Test if the assumption was correct:
$$\frac{ 1.9 \times 10^{-3} }{ 0.175 } \times 100\% = 1.1 \%$$
7. Thus, it is correct to say that $x = 1.9 \times 10^{-3} $
8. $[OH^-] = x = 1.9 \times 10^{-3} $
9. Calculate the pH:
$$[H_3O^+] = \frac{1.0 \times 10^{-14}}{[OH^-]} = \frac{1.0 \times 10^{-14}}{ 1.9 \times 10^{-3} } = 5.3 \times 10^{-12} \space M$$
$$pH = -log[H_3O^+] = -log( 5.3 \times 10^{-12} ) = 11.28 $$
