Answer
(a) pH = 9.55
(b) pH = 9.50
Work Step by Step
(a)
- Converting the amount in moles to concentration ($NH_4Cl$):
** $5.00 \times 10^2 $ mL = $500 mL$ = 0.500 L
$C(M) = \frac{Amount(moles)}{volume(L)} = \frac{0.125}{0.500} = 0.250M$
1. Since $NH{_4}^+$ 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}$
$K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$
$K_a = 5.56\times 10^{- 10}$
2. Calculate the pKa Value
$pKa = -log(Ka)$
$pKa = -log( 5.6 \times 10^{- 10})$
$pKa = 9.25$
3. Check if the ratio is between 0.1 and 10:
- $\frac{[Base]}{[Acid]} = \frac{0.5}{0.25}$
- 2: It is.
4. Check if the compounds exceed the $K_a$ by 100 times or more:
- $ \frac{0.5}{5.56 \times 10^{-10}} = 8.99\times 10^{8}$
- $ \frac{0.25}{5.56 \times 10^{-10}} = 4.5\times 10^{8}$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 9.25 + log(\frac{0.5}{0.25})$
$pH = 9.25 + log(2)$
$pH = 9.25 + 0.301$
$pH = 9.55$
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(b)
When we add $HCl$, this reaction will occur:
$HCl + NH_3 -- \gt NH_4^+ + Cl^-$
**$[HCl] = \frac{0.0100}{0.500} = 0.0200M$
Therefore, the concentration of the base will decrease, and the acid's concentration will be increased. Since the coefficients are 1, these are the new concentrations:
$[Acid] = 0.25 + 0.0200 = 0.27 M$
$[Base] = 0.50 - 0.0200 = 0.48 M$
$pH = 9.25 + log\frac{0.48}{0.27} = 9.25 + 0.25 = 9.50$
