Answer
The equilibrium constant for this reaction is $4.0 \times 10^{-2}$
Work Step by Step
1. According to the Table 15.3, these reactions have the following equilibrium constant:
$$CH_3COOH \leftrightharpoons CH_3COO^- + H^+$$
$K' = 1.8 \times 10^{-5}$
$$HNO_2 \leftrightharpoons H^+ + NO{_2}^-$$
$K'' = 4.5 \times 10^{-4}$
2. Reverse the second reaction:
$$H^++N{O_2}^-\leftrightharpoons HNO_2$$
$K''' = \frac{1}{K''} = 2.2 \times 10^3$
3. Add equation 1 and 3:
$$CH_3COOH + H^++N{O_2}^- \leftrightharpoons CH_3COO^- + H^+ + N{O_2}^-$$
$$CH_3COOH +N{O_2}^- \leftrightharpoons CH_3COO^- + N{O_2}^-$$
$$K = K' \times K''' = 1.8 \times 10^{-5} \times 2.2 \times 10^3 = 4.0 \times 10^{-2}$$
