Answer
The pH of this solution is equal to $12.10$.
Work Step by Step
1000ml = 1L
20ml = 0.02 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HClO_4) * V(HClO_4) = 0.125* 0.02 = 2.5 \times 10^{-3}$ moles
$C(KOH) * V(KOH) = 0.15* 0.02 = 3 \times 10^{-3}$ moles
2. Write the acid-base reaction:
$HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$
- Total volume: 0.02 + 0.02 = 0.04L
3. Since the acid is the limiting reactant, only $ 0.0025$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HClO_4] = 0.0025 - 0.0025 = 0M$.
$[KOH] = 0.003 - 0.0025 = 5 \times 10^{-4}$ mol
Concentration: $\frac{5 \times 10^{-4}}{ 0.04} = 0.0125M$
- The only significant electrolyte in the solution is $KOH$, which is a strong base, so:
$[OH^-] = [KOH] = 0.0125M$
$pOH = -log[OH^-]$
$pOH = -log( 0.0125)$
$pOH = 1.903$
$pH + pOH = 14$
$pH + 1.903 = 14$
$pH = 12.097$
