Chemistry 10th Edition

- Since we want a $pH \approx 9$, we will have to choose a combination where the $pK_a \approx 9$. (a) 1. Calculate the pKa value $(CH_3COOH)$ $pKa = -log(Ka)$ $pKa = -log( 1.8 \times 10^{- 5})$ $pKa = 4.75$ (b) - $HCl$: Strong acid: $pK_a < 0$ (c) 2. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its kb by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.6\times 10^{- 10}$ 3. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.6 \times 10^{- 10})$ $pKa = 9.25$ Therefore, (c) is the best combination.