Answer
(c) is the best combination.
Work Step by Step
- 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.
