Answer
$pH = 9.25$
Correct answer: $(b)$
Work Step by Step
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
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa = 9.25$
1000ml = 1L
200ml = 0.2 L
100ml = 0.1 L
3. Find the numbers of moles:
$C(NH{_4}^+) * V(NH{_4}^+) = 0.1*0.2 = 0.02$ moles
$C(NH_3) * V(NH_3) = 0.2*0.1 = 0.02$ moles
4. Calculate the total volume:
- Total volume: 0.2 + 0.1 = 0.3L
5. Calculate the final concentrations:
$[NH{_4}^+] : \frac{0.02}{0.3} = 0.067M $
$[NH_3] : \frac{0.02}{0.3} = 0.067M $
6. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa = 9.25$
7. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.067}{0.067}$
- 1: It is.
8. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.067}{5.6 \times 10^{-10}} = 1.2\times 10^{8}$
- $ \frac{0.067}{5.6 \times 10^{-10}} = 1.2\times 10^{8}$
9. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.067}{0.067})$
$pH = 9.25 + log(1)$
$pH = 9.25 + 0$
$pH = 9.25$
