Answer
$pH = 4.38$
Work Step by Step
1000ml = 1L
50ml = 0.05 L
30ml = 0.03 L
1. Find the numbers of moles:
$C(C_6H_5COOH) * V(C_6H_5COOH) = 0.015* 0.05 = 7.5 \times 10^{-4}$ moles
$C(KOH) * V(KOH) = 0.015* 0.03 = 4.5 \times 10^{-4}$ moles
2. Write the acid-base reaction:
$C_6H_5COOH(aq) + KOH(aq) -- \gt KC_6H_5COO(aq) + H_2O(l)$
- Total volume: 0.05 + 0.03 = 0.08L
3. Since the base is the limiting reactant, only $ 0.00045$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[C_6H_5COOH] = 0.00075 - 0.00045 = 3 \times 10^{-4}$ moles.
Concentration: $\frac{3 \times 10^{-4}}{ 0.08} = 3.8 \times 10^{-3}M$
$[KOH] = 0.00045 - 0.00045 = 0$
$[KC_6H_5COO] = 0 + 0.00045 = 0.00045$ moles.
Concentration: $\frac{ 0.00045}{ 0.08} = 5.6 \times 10^{-3}M$
4. Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 6.3 \times 10^{- 5})$
$pKa = 4.2$
5. Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.2 + log(\frac{5.6 \times 10^{-3}}{3.8 \times 10^{-3}})$
$pH = 4.2 + log(1.5)$
$pH = 4.2 + 0.176$
$pH = 4.38$
