Answer
(a) $[OH^-] = 3.0 \times 10^{-5}M$ and $pH = 9.48$
(b) $[OH^-] = 6.8 \times 10^{-6}M$ and $pH = 8.83$
Work Step by Step
(a)
1. 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}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.556\times 10^{- 10}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.556 \times 10^{- 10})$
$pKa = 9.255$
3. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.25}{0.15}$
- 1.667: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.25}{ 5.56 \times 10^{-10}} = 4.5\times 10^{8}$
- $ \frac{0.15}{ 5.56 \times 10^{-10}} = 2.7\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.255 + log(\frac{0.25}{0.15})$
$pH = 9.255 + 0.2218$
$pH = 9.477$
6. Calculate the hydroxide concentration:
pH + pOH = 14
9.477 + pOH = 14
pOH = 4.523
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 4.523}$
$[OH^-] = 2.999 \times 10^{- 5}$
(b)
1 and 2. We've already calculated the $pK_a$ value for $N{H_4}^+$
3. Che
** $(NH_4)_2SO_4 = 0.20M$; therefore: $N{H_4}^+ = 0.40M$
ck if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.15}{0.4}$
- 0.375: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.15}{5.556 \times 10^{-10}} = 2.7\times 10^{8}$
- $ \frac{0.4}{5.556 \times 10^{-10}} = 7.2\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.255 + log(\frac{0.15}{0.4})$
$pH = 9.255 + -0.4259$
$pH = 8.829$
6. Calculate the $[OH^-]$
pH + pOH = 14
8.829 + pOH = 14
pOH = 5.171
$[OH^-] = 10^{-pOH}$
$[OH^-] = 10^{- 5.171}$
$[OH^-] = 6.745 \times 10^{- 6}$
