General Chemistry: Principles and Modern Applications (10th Edition)

(a) $pH = 11.85$ (b) $pH = 1.426$
(a) 1000ml = 1L 15ml = 0.015 L 20ml = 0.02 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.35* 0.015 = 5.25 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.275* 0.02 = 5.5 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$ - Total volume: 0.015 + 0.02 = 0.035L 3. Since the acid is the limiting reactant, only $0.00525$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.00525 - 0.00525 = 0M$. $[KOH] = 0.0055 - 0.00525 = 2.5 \times 10^{-4}$ mol Concentration: $\frac{2.5 \times 10^{-4}}{ 0.035} = 7.1 \times 10^{-3}M$ - The only significant electrolyte in the solution is $KOH$, which is a strong base, so: $[OH^-] = [KOH] = 7.1 \times 10^{-3}M$ 4. Calculate the pOH: $pOH = -log[OH^-]$ $pOH = -log( 7.1 \times 10^{- 3})$ $pOH = 2.15$ 5. Find the pH: $pH + pOH = 14$ $pH + 2.15 = 14$ $pH = 11.85$ ------ (b) 1000ml = 1L 20ml = 0.02 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.35* 0.02 = 7 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.275* 0.02 = 5.5 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$ - Total volume: 0.02 + 0.02 = 0.04L 3. Since the base is the limiting reactant, only $0.0055$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.007 - 0.0055 = 1.5 \times 10^{-3}$ moles. Concentration: $\frac{1.5 \times 10^{-3}}{ 0.04} = 0.0375M$ $[KOH] = 0.0055 - 0.0055 = 0$ moles - Since : $HCl$ is a strong acid: $[HCl] = [H_3O^+] = 0.0375M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.0375)$ $pH = 1.426$