## Chemistry 9th Edition

a. $pH = 0.699$ b. $pH = 0.854$ c. $pH = 1.301$ d. $pH = 7.00$ e. $pH = 12.155$
(a) - Since : $HClO_4$ is a strong acid, and we have a pure solution: $[HClO_4] = [H_3O^+] = 0.2M$ 1. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.2)$ $pH = 0.699$ (b) 1000ml = 1L 40ml = 0.04 L 10ml = 0.01 L 1. Find the numbers of moles: $C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.01 = 1 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$ - Total volume: 0.04 + 0.01 = 0.05L 3. Since the base is the limiting reactant, only $0.001$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HClO_4] = 0.008 - 0.001 = 7 \times 10^{-3}$ moles. Concentration: $\frac{7 \times 10^{-3}}{ 0.05} = 0.14M$ $[KOH] = 0.001 - 0.001 = 0$ moles - Since : $HClO_4$ is a strong acid: $[HClO_4] = [H_3O^+] = 0.14M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.14)$ $pH = 0.8539$ (c) 1000ml = 1L 40ml = 0.04 L 40ml = 0.04 L 1. Find the numbers of moles: $C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.04 = 4 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$ - Total volume: 0.04 + 0.04 = 0.08L 3. Since the base is the limiting reactant, only $0.004$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HClO_4] = 0.008 - 0.004 = 4 \times 10^{-3}$ moles. Concentration: $\frac{4 \times 10^{-3}}{ 0.08} = 0.05M$ $[KOH] = 0.004 - 0.004 = 0$ moles - Since : $HClO_4$ is a strong acid: $[HClO_4] = [H_3O^+] = 0.05M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.05)$ $pH = 1.301$ (d) 1000ml = 1L 40ml = 0.04 L 80ml = 0.08 L 1. Find the numbers of moles: $C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.08 = 8 \times 10^{-3}$ moles 2. When the number of moles is equal, the reactants are totally consumed: $HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$ - Total volume: 0.04 + 0.08 = 0.12L 3. So, those are the final concentrations: $[HClO_4] = 0.008 - 0.008 = 0$ mol. $[KOH] = 0.008 - 0.008 = 0$ mol - Therefore, the solution doesn't have any significant electrolyte: pH = 7 (Neutral) (e) 1000ml = 1L 40ml = 0.04 L 100ml = 0.1 L 1. Find the numbers of moles: $C(HClO_4) * V(HClO_4) = 0.2* 0.04 = 8 \times 10^{-3}$ moles $C(KOH) * V(KOH) = 0.1* 0.1 = 0.01$ moles 2. Write the acid-base reaction: $HClO_4(aq) + KOH(aq) -- \gt KClO_4(aq) + H_2O(l)$ - Total volume: 0.04 + 0.1 = 0.14L 3. Since the acid is the limiting reactant, only $0.008$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HClO_4] = 0.008 - 0.008 = 0M$. $[KOH] = 0.01 - 0.008 = 2 \times 10^{-3}$ mol Concentration: $\frac{2 \times 10^{-3}}{ 0.14} = 0.01429M$ - The only significant electrolyte in the solution is $KOH$, which is a strong base, so: $[OH^-] = [KOH] = 0.01429M$ $pOH = -log[OH^-]$ $pOH = -log( 0.01429)$ $pOH = 1.845$ $pH + pOH = 14$ $pH + 1.845 = 14$ $pH = 12.155$