## Chemistry and Chemical Reactivity (9th Edition)

$pH = 9.25$ Correct answer: $(b)$
1. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.6\times 10^{- 10}$ 2. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.6 \times 10^{- 10})$ $pKa = 9.25$ 1000ml = 1L 200ml = 0.2 L 100ml = 0.1 L 3. Find the numbers of moles: $C(NH{_4}^+) * V(NH{_4}^+) = 0.1*0.2 = 0.02$ moles $C(NH_3) * V(NH_3) = 0.2*0.1 = 0.02$ moles 4. Calculate the total volume: - Total volume: 0.2 + 0.1 = 0.3L 5. Calculate the final concentrations: $[NH{_4}^+] : \frac{0.02}{0.3} = 0.067M$ $[NH_3] : \frac{0.02}{0.3} = 0.067M$ 6. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.6 \times 10^{- 10})$ $pKa = 9.25$ 7. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.067}{0.067}$ - 1: It is. 8. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.067}{5.6 \times 10^{-10}} = 1.2\times 10^{8}$ - $\frac{0.067}{5.6 \times 10^{-10}} = 1.2\times 10^{8}$ 9. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.25 + log(\frac{0.067}{0.067})$ $pH = 9.25 + log(1)$ $pH = 9.25 + 0$ $pH = 9.25$