## Chemistry (7th Edition)

(a) $pH = 0.996$ (b) $pH = 1.602$ (c) $pH = 12.57$
(a) 1000ml = 1L 25ml = 0.025 L 3ml = $3 \times 10^{-3} L$ 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.125* 0.025 = 3.13 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.003 = 3 \times 10^{-4}$ moles 2. Write the acid-base reaction: $HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$ - Total volume: 0.025 + 0.003 = 0.028L 3. Since the base is the limiting reactant, only $0.0003$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.003125 - 0.0003 = 2.83 \times 10^{-3}$ moles. Concentration: $\frac{2.83 \times 10^{-3}}{ 0.028} = 0.101M$ $[KOH] = 0.0003 - 0.0003 = 0$ moles - Since $HCl$ is a strong acid: $[HCl] = [H_3O^+] = 0.101M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.101)$ $pH = 0.9961$ (b) 1000ml = 1L 25ml = 0.025 L 20ml = 0.02 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.125* 0.025 = 3.13 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.02 = 2 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$ - Total volume: 0.025 + 0.02 = 0.045L 3. Since the base is the limiting reactant, only $0.002$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.003125 - 0.002 = 1.12 \times 10^{-3}$ moles. Concentration: $\frac{1.12 \times 10^{-3}}{ 0.045} = 0.025M$ $[KOH] = 0.002 - 0.002 = 0$ moles - Since $HCl$ is a strong acid: $[HCl] = [H_3O^+] = 0.025M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.025)$ $pH = 1.602$ (c) 1000ml = 1L 25ml = 0.025 L 65ml = 0.065 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.125* 0.025 = 3.13 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.065 = 6.5 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + KOH(aq) -- \gt KCl(aq) + H_2O(l)$ - Total volume: 0.025 + 0.065 = 0.09L 3. Since the acid is the limiting reactant, only $0.003125$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.003125 - 0.003125 = 0M$. $[KOH] = 0.0065 - 0.003125 = 3.38 \times 10^{-3}$ mol Concentration: $\frac{3.38 \times 10^{-3}}{ 0.09} = 0.0375M$ - The only significant electrolyte in the solution is $KOH$, which is a strong base, so: $[OH^-] = [KOH] = 0.0375M$ $pOH = -log[OH^-]$ $pOH = -log( 0.0375)$ $pOH = 1.43$ $pH + pOH = 14$ $pH + 1.43 = 14$ $pH = 12.57$