## Chemistry: The Central Science (13th Edition)

The pH of this solution is equal to $12.10$.
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: $[OH^-] = [KOH] = 0.0125M$ $pOH = -log[OH^-]$ $pOH = -log( 0.0125)$ $pOH = 1.903$ $pH + pOH = 14$ $pH + 1.903 = 14$ $pH = 12.097$