## Chemistry and Chemical Reactivity (9th Edition)

$pH = 9.66$
1000ml = 1L 25 ml = 0.025 L 25 ml = 0.025 L 1. Find the numbers of moles: $C(HCl) * V(HCl) = 0.12* 0.025 = 3 \times 10^{-3}$ moles $C(NH_3) * V(NH_3) = 0.43* 0.025 = 0.011$ moles 2. Write the acid-base reaction: $HCl(aq) + NH_3(aq) -- \gt N{H_4}^+(aq) + Cl^-(aq)$ - Total volume: 0.025 + 0.025 = 0.05L 3. Since the acid is the limiting reactant, only $0.003$ mol of the compounds will react. Therefore: Concentration (M) = $\frac{n(mol)}{Volume(L)}$ $[HCl] = 0.003 - 0.003 = 0$ mol. $[NH_3] = 0.01075 - 0.003 = 7.8 \times 10^{-3} mol$ Concentration: $\frac{7.8 \times 10^{-3}}{ 0.05} = 0.16M$ $[N{H_4}^+] = 0 + 0.003 = 0.003$ moles. Concentration: $\frac{ 0.003}{ 0.05} = 0.06M$ 4. Since $N{H_4}^+$ is the conjugate acid of $NH_3$ , we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.6\times 10^{- 10}$ 5. Calculate the $pK_a$ for the acid $pKa = -log(Ka)$ $pKa = -log( 5.6 \times 10^{- 10})$ $pKa = 9.25$ 6. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.25 + log(\frac{0.16}{0.06})$ $pH = 9.25 + log(2.6)$ $pH = 9.25 + 0.41$ $pH = 9.66$