Answer
(a) $[H_3O^+] = 2.138 \times 10^{- 3}M$
(b) $Ka = 3.554\times 10^{- 4}$
Work Step by Step
1. Drawing the equilibrium (ICE) table, we get these concentrations at equilibrium:** The image is in the end of this answer.
-$[H_3O^+] = [OCN^-] = x$
-$[HOCN] = [HOCN]_{initial} - x$
2. Calculate the hydronium concentration:
$[H_3O^+] = 10^{-pH}$
$[H_3O^+] = 10^{- 2.67}$
$[H_3O^+] = 2.138 \times 10^{- 3}$
3. Write the Ka equation, and find its value:
$Ka = \frac{[H_3O^+][OCN^-]}{ [HOCN]}$
$Ka = \frac{x^2}{[InitialHOCN] - x}$
$Ka = \frac{( 2.138\times 10^{- 3})^2}{ 0.015- 2.138\times 10^{- 3}}$
$Ka = \frac{ 4.571\times 10^{- 6}}{ 0.01286}$
$Ka = 3.554\times 10^{- 4}$
