Answer
$pH = 11.00$
Work Step by Step
- Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.1* 0.05 = 5 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.1* 0.051 = 5.1 \times 10^{-3}$ moles
Write the acid-base reaction:
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
- Total volume: 0.05 + 0.051 = 0.101L
Since the acid is the limiting reactant, only $ 0.005$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.005 - 0.005 = 0M$.
$[NaOH] = 0.0051 - 0.005 = 1.0 \times 10^{-4}$ mol
Concentration: $\frac{1.0 \times 10^{-4}}{ 0.101} = 9.901 \times 10^{-4}M$
$[NaCH_3COO] = 0 + 0.005 = 0.005$ moles.
Concentration: $\frac{ 0.005}{ 0.101} = 0.0495M$
- We have a strong and a weak base. We can ignore the weak one, and we calculate the pH based only on the strong base concentration:
$[OH^-] = [NaOH]$
$pOH = -log[OH^-]$
$pOH = -log( 9.901 \times 10^{- 4})$
$pOH = 3.004$
$pH + pOH = 14$
$pH + 3.004 = 14$
$pH = 11.00$