Answer
The pH of this solution is going to be equal to $11.03$
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
$C(NaOH) * V(NaOH) = 0.15* 0.0355 = 5.324 \times 10^{-3}$ moles
Write the acid-base reaction:
$CH_3COOH(aq) + NaOH(aq) -- \gt NaCH_3COO(aq) + H_2O(l)$
- Total volume: 0.035 + 0.0355 = 0.0705L
Since the acid is the limiting reactant, only $ 0.00525$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[CH_3COOH] = 0.00525 - 0.00525 = 0M$.
$[NaOH] = 0.005325 - 0.00525 = 7.5 \times 10^{-5}$ mol
Concentration: $\frac{7.5 \times 10^{-5}}{ 0.0705} = 1.064 \times 10^{-3}M$
$[NaCH_3COO] = 0 + 0.00525 = 0.00525$ moles.
Concentration: $\frac{ 0.00525}{ 0.0705} = 0.07446M$
- 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:
$[OH^-] = [NaOH]$
$pOH = -log[OH^-]$
$pOH = -log( 1.064 \times 10^{- 3})$
$pOH = 2.973$
$pH + pOH = 14$
$pH + 2.973 = 14$
$pH = 11.027$
