Answer
(b)
Work Step by Step
1. First, determine the $pK_a$ for each of the acids:
(a) $HCl$: Strong acid: $pK_a < 0$
(b)
1. Since $N{H_4}^+$ is the conjugate acid of $NH_3$, we can calculate its $K_a$ by using this equation:
$K_b * K_a = K_w = 10^{-14}$
$ 1.8\times 10^{- 5} * K_a = 10^{-14}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.6\times 10^{- 10}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa (N{H_4}^+) = 9.25$
(c)
1. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa (CH_3CO_2H) = 4.74$
2. Choose the one with the closest $pK_a$ to the desired pH (9);
That would be $N{H_4}^+$: Answer = (b)
