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

(a) $pH = 8.78$ (b) $pH = 1.66$
1000ml = 1L 10ml = 0.01 L 20ml = 0.02 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.475* 0.01 = 4.75 \times 10^{-3}$ moles $C(NH_3) * V(NH_3) = 0.318* 0.02 = 6.36 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NH_3(aq) -- \gt N{H_4}^+(aq) + Cl^-(aq)$ - Total volume: 0.01 + 0.02 = 0.03L 3. Since the acid is the limiting reactant, only $0.00475$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.00475 - 0.00475 = 0$ mol. $[NH_3] = 0.00636 - 0.00475 = 1.61 \times 10^{-3} mol$ Concentration: $\frac{1.61 \times 10^{-3}}{ 0.03} = 0.0537M$ $[N{H_4}^+] = 0 + 0.00475 = 0.00475$ moles. Concentration: $\frac{ 0.00475}{ 0.03} = 0.158M$ 4. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.56\times 10^{- 10}$ 5. Calculate the $pK_a$ for the acid $pKa = -log(Ka)$ $pKa = -log( 5.56 \times 10^{- 10})$ $pKa = 9.25$ 6. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.25 + log(\frac{0.0537}{0.158})$ $pH = 9.25 + log(0.339)$ $pH = 9.25 + (-0.47)$ $pH = 8.78$ ----- (b) 1000ml = 1L 15ml = 0.015 L 20ml = 0.02 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.475* 0.015 = 7.12 \times 10^{-3}$ moles $C(NH_3) * V(NH_3) = 0.318* 0.02 = 6.36 \times 10^{-3}$ moles 2. Write the acid-base reaction: $HCl(aq) + NH_3(aq) -- \gt N{H_4}^+(aq) + Cl^-(aq)$ - Total volume: 0.015 + 0.02 = 0.035L Since the base is the limiting reactant, only $0.00636$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[NH_3] = 0.00636 - 0.00636 = 0M$. $[HCl] = 0.007125 - 0.00636 = 7.65 \times 10^{-4}$ mol Concentration: $\frac{7.65 \times 10^{-4}}{ 0.035} = 0.0219M$ $[N{H_4}^+] = 0 + 0.00636 = 0.00636$ moles. Concentration: $\frac{ 0.00636}{ 0.035} = 0.182M$ - We have a strong and a weak acid. We can ignore the weak one (because it has a very low $K_a$), and calculate the pH based only on the strong acid concentration: $[H_3O^+] = [HCl] = 0.0219M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 0.0219)$ $pH = 1.66$