Answer
$pH = 12.15$
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
$C(NaOH) * V(NaOH) = 0.1* 0.04 = 4 \times 10^{-3}$ moles
Write the acid-base reaction:
$C_6H_5COOH(aq) + NaOH(aq) -- \gt NaC_6H_5COO(aq) + H_2O(l)$
- Total volume: 0.03 + 0.04 = 0.07L
Since the acid is the limiting reactant, only $ 0.003$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[C_6H_5COOH] = 0.003 - 0.003 = 0M$.
$[NaOH] = 0.004 - 0.003 = 0.001$ mol
Concentration: $\frac{ 0.001}{ 0.07} = 0.01429M$
$[NaC_6H_5COO] = 0 + 0.003 = 0.003$ moles.
Concentration: $\frac{ 0.003}{ 0.07} = 0.04286M$
- We have a strong and a weak base. We can ignore the weak one, and calculate the pH based only on the strong base concentration:
$[OH^-] = [NaOH]$
$pOH = -log[OH^-]$
$pOH = -log( 0.01429)$
$pOH = 1.85$
$pH + pOH = 14$
$pH + 1.845 = 14$
$pH = 12.15$
