Answer
The pH of this solution is going to be equal to $6.58$.
Work Step by Step
- Find the numbers of moles:
$C(CH_3COOH) * V(CH_3COOH) = 0.15* 0.035 = 5.25 \times 10^{-3}$ moles
$C(NaOH) * V(NaOH) = 0.15* 0.0345 = 5.175 \times 10^{-3}$ moles
Write the acid-base reaction:
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
- Total volume: 0.035 + 0.0345 = 0.0695L
Since the base is the limiting reactant, only $ 0.005175$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.00525 - 0.005175 = 7.5 \times 10^{-5}$ moles.
Concentration: $\frac{7.5 \times 10^{-5}}{ 0.0695} = 1.079 \times 10^{-3}M$
$[NaOH] = 0.005175 - 0.005175 = 0$
$[NaCH_3COO] = 0 + 0.005175 = 0.005175$ moles.
Concentration: $\frac{ 0.005175}{ 0.0695} = 0.07445M$
- Calculate the $pK_a$ for the acid
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.745$
- Using the Henderson–Hasselbalch equation:
$pH = pKa + log(\frac{[Base]}{[Acid]})$
$pH = 4.745 + log(\frac{0.07445}{1.079 \times 10^{-3}})$
$pH = 4.745 + log(69)$
$pH = 4.745 + 1.839$
$pH = 6.584$
