Chemistry (7th Edition)

Published by Pearson

Chapter 16 - Applications of Aqueous Equilibria - Section Problems - Page 710: 87

Answer

(a) $pH = 12.99$ (b) $pH = 12.31$ (c) $pH = 1.95$

Work Step by Step

(a) 1000ml = 1L 5ml =$5 \times 10^{-3} L$ 50ml = 0.05 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.075* 0.005 = 3.75 \times 10^{-4}$ moles $C(NaOH) * V(NaOH) = 0.116* 0.05 = 5.8 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ - Total volume: 0.005 + 0.05 = 0.055L 3. Since the acid is the limiting reactant, only $0.000375$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.000375 - 0.000375 = 0M$. $[NaOH] = 0.0058 - 0.000375 = 5.43 \times 10^{-3}$ mol Concentration: $\frac{5.43 \times 10^{-3}}{ 0.055} = 0.0985M$ - The only significant electrolyte in the solution is $NaOH$, which is a strong base, so: $[OH^-] = [NaOH] = 0.0985M$ $pOH = -log[OH^-]$ $pOH = -log( 0.0986)$ $pOH = 1.01$ $pH + pOH = 14$ $pH + 1.01 = 14$ $pH = 12.99$ (b) 1000ml = 1L 50ml = 0.05 L 50ml = 0.05 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.075* 0.05 = 3.75 \times 10^{-3}$ moles $C(NaOH) * V(NaOH) = 0.116* 0.05 = 5.8 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ - Total volume: 0.05 + 0.05 = 0.1L 3. Since the acid is the limiting reactant, only $0.00375$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.00375 - 0.00375 = 0M$. $[NaOH] = 0.0058 - 0.00375 = 2.04 \times 10^{-3}$ mol Concentration: $\frac{2.04 \times 10^{-3}}{ 0.1} = 0.0205M$ - The only significant electrolyte in the solution is $NaOH$, which is a strong base, so: $[OH^-] = [NaOH] = 0.0205M$ $pOH = -log[OH^-]$ $pOH = -log( 0.0205)$ $pOH = 1.69$ $pH + pOH = 14$ $pH + 1.69 = 14$ $pH = 12.31$ (c) 1000ml = 1L 50ml = 0.05 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.075* 0.1 = 7.5 \times 10^{-3}$ moles $C(NaOH) * V(NaOH) = 0.116* 0.05 = 5.8 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NaOH(aq) -- \gt NaCl(aq) + H_2O(l)$ - Total volume: 0.1 + 0.05 = 0.15L 3. Since the base is the limiting reactant, only $0.0058$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.0075 - 0.0058 = 1.6 \times 10^{-3}$ moles. Concentration: $\frac{1.6 \times 10^{-3}}{ 0.15} = 0.0112M$ $[NaOH] = 0.0058 - 0.0058 = 0$ moles - Since $HCl$ is a strong acid: $[HCl] = [H_3O^+] = 0.0112M$ 5. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.0113)$ $pH = 1.95$

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