## Chemistry and Chemical Reactivity (9th Edition)

(a) - Converting the amount in moles to concentration ($NH_4Cl$): ** $5.00 \times 10^2$ mL = $500 mL$ = 0.500 L $C(M) = \frac{Amount(moles)}{volume(L)} = \frac{0.125}{0.500} = 0.250M$ 1. Since $NH{_4}^+$ is the conjugate acid of $NH_3$, we can calculate its $K_a$ by using this equation: $K_b * K_a = K_w = 10^{-14}$ $1.8\times 10^{- 5} * K_a = 10^{-14}$ $K_a = \frac{10^{-14}}{ 1.8\times 10^{- 5}}$ $K_a = 5.56\times 10^{- 10}$ 2. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 5.6 \times 10^{- 10})$ $pKa = 9.25$ 3. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.5}{0.25}$ - 2: It is. 4. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.5}{5.56 \times 10^{-10}} = 8.99\times 10^{8}$ - $\frac{0.25}{5.56 \times 10^{-10}} = 4.5\times 10^{8}$ 5. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 9.25 + log(\frac{0.5}{0.25})$ $pH = 9.25 + log(2)$ $pH = 9.25 + 0.301$ $pH = 9.55$ ------ (b) When we add $HCl$, this reaction will occur: $HCl + NH_3 -- \gt NH_4^+ + Cl^-$ **$[HCl] = \frac{0.0100}{0.500} = 0.0200M$ Therefore, the concentration of the base will decrease, and the acid's concentration will be increased. Since the coefficients are 1, these are the new concentrations: $[Acid] = 0.25 + 0.0200 = 0.27 M$ $[Base] = 0.50 - 0.0200 = 0.48 M$ $pH = 9.25 + log\frac{0.48}{0.27} = 9.25 + 0.25 = 9.50$