## Chemistry 10th Edition

(a) $[OH^-] = 3.0 \times 10^{-5}M$ and $pH = 9.48$ (b) $[OH^-] = 6.8 \times 10^{-6}M$ and $pH = 8.83$
(a) 1. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its kb by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.556\times 10^{- 10}$ 2. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.556 \times 10^{- 10})$ $pKa = 9.255$ 3. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.25}{0.15}$ - 1.667: It is. 4. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.25}{ 5.56 \times 10^{-10}} = 4.5\times 10^{8}$ - $\frac{0.15}{ 5.56 \times 10^{-10}} = 2.7\times 10^{8}$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.255 + log(\frac{0.25}{0.15})$ $pH = 9.255 + 0.2218$ $pH = 9.477$ 6. Calculate the hydroxide concentration: pH + pOH = 14 9.477 + pOH = 14 pOH = 4.523 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 4.523}$ $[OH^-] = 2.999 \times 10^{- 5}$ (b) 1 and 2. We've already calculated the $pK_a$ value for $N{H_4}^+$ 3. Che ** $(NH_4)_2SO_4 = 0.20M$; therefore: $N{H_4}^+ = 0.40M$ ck if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.15}{0.4}$ - 0.375: It is. 4. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.15}{5.556 \times 10^{-10}} = 2.7\times 10^{8}$ - $\frac{0.4}{5.556 \times 10^{-10}} = 7.2\times 10^{8}$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.255 + log(\frac{0.15}{0.4})$ $pH = 9.255 + -0.4259$ $pH = 8.829$ 6. Calculate the $[OH^-]$ pH + pOH = 14 8.829 + pOH = 14 pOH = 5.171 $[OH^-] = 10^{-pOH}$ $[OH^-] = 10^{- 5.171}$ $[OH^-] = 6.745 \times 10^{- 6}$