Answer
The pH of this solution is equal to: $11.46$
Work Step by Step
1000ml = 1L
23ml = 0.023 L
20ml = 0.02 L
1. Find the numbers of moles:
$C(HClO_4) * V(HClO_4) = 0.125* 0.023 = 2.875 \times 10^{-3}$ 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: 0.023 + 0.02 = 0.043L
3. Since the acid is the limiting reactant, only $ 0.002875$ mol of the compounds will react.
Therefore:
Concentration (M) = $\frac{n(mol)}{Volume(L)}$
$[HClO_4] = 0.002875 - 0.002875 = 0M$.
$[KOH] = 0.003 - 0.002875 = 1.25 \times 10^{-4}$ mol
Concentration: $\frac{1.25 \times 10^{-4}}{ 0.043} = 2.907 \times 10^{-3}M$
- The only significant electrolyte in the solution is $KOH$, which is a strong base, so:
$[OH^-] = [KOH] = 2.907 \times 10^{-3}M$
$pOH = -log[OH^-]$
$pOH = -log( 2.907 \times 10^{- 3})$
$pOH = 2.537$
$pH + pOH = 14$
$pH + 2.537 = 14$
$pH = 11.463$
