Answer
$pH = 3.62$
Work Step by Step
- Find the numbers of moles:
$C(C_6H_5COOH) * V(C_6H_5COOH) = 0.1* 0.03 = 3 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.1* 0.01 = 1 \times 10^{-3}$ moles
Write the acid-base reaction:
$C_6H_5COOH(aq) + NaOH(aq) \ -- \gt NaC_6H_5COO(aq) + H_2O(l)$
- Total volume: 0.03 + 0.01 = 0.04L
Since the base is the limiting reactant, only $ 0.001$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[C_6H_5COOH] = 0.003 - 0.001 = 0.002$ moles.
Concentration: $\frac{ 0.002}{ 0.04} = 0.05M$
$[NaOH] = 0.001 - 0.001 = 0$
$[NaC_6H_5COO] = 0 + 0.001 = 0.001$ moles.
Concentration: $\frac{ 0.001}{ 0.04} = 0.025M$
- Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 1.2 \times 10^{- 4})$
$pKa = 3.921$
- Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 3.921 + log(\frac{ 0.025}{ 0.05})$
$pH = 3.921 + log(0.5)$
$pH = 3.921 + -0.301$
$pH = 3.62$
