## Chemistry and Chemical Reactivity (9th Edition)

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)