## Chemistry and Chemical Reactivity (9th Edition)

(a) $pH = 4.95$ (b) $pH = 5.05$
1. Calculate the molar mass $(NaOH)$: 22.99* 1 + 16* 1 + 1.01* 1 = 40g/mol 2. Calculate the number of moles $(NaOH)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 0.082}{ 40}$ $n(moles) = 2.1\times 10^{- 3}$ 3. Find the concentration in mol/L $(NaOH)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 2.1\times 10^{- 3}}{ 0.1}$ $C(mol/L) = 0.021$ 4. Calculate the molar mass $(NaCH_3CO_2)$: 22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 16* 2 = 82.04g/mol 5. Calculate the number of moles $(NaCH_3CO_2)$ $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{ 4.95}{ 82.04}$ $n(moles) = 0.06$ 6. Find the concentration in mol/L $(NaCH_3CO_2)$: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.06}{ 0.25}$ $C(mol/L) = 0.24$ 7. Drawing the ICE table we get these concentrations at the equilibrium: $CH_3CO_2H(aq) + H_2O(l) \lt -- \gt CH_3C{O_2}^-(aq) + H_3O^+(aq)$ Remember: Reactants at equilibrium = Initial Concentration - x And Products = Initial Concentration + x $[CH_3CO_2H] = 0.15 M - x$ $[CH_3C{O_2}^-] = 0.24M + x$ $[H_3O^+] = 0 + x$ 8. Calculate 'x' using the $K_a$ expression. $1.8\times 10^{- 5} = \frac{[CH_3C{O_2}^-][H_3O^+]}{[CH_3CO_2H]}$ $1.8\times 10^{- 5} = \frac{( 0.24 + x )* x}{ 0.15 - x}$ Considering 'x' has a very small value. $1.8\times 10^{- 5} = \frac{ 0.24 * x}{ 0.15}$ $1.8\times 10^{- 5} = 1.6x$ $\frac{ 1.8\times 10^{- 5}}{ 1.6} = x$ $x = 1.13\times 10^{- 5}$ Percent dissociation: $\frac{ 1.13\times 10^{- 5}}{ 0.15} \times 100\% = 7.5\times 10^{- 3}\%$ x = $[H_3O^+]$ $[CH_3CO_2H] = 0.15 M - x = 0.15 M - 1.13 \times 10^{-5}M \approx 0.15M$ $[CH_3C{O_2}^-] = 0.24M + x = 0.15 M + 1.13 \times 10^{-5}M \approx 0.24M$ $[H_3O^+] = 0 + x = 1.13 \times 10^{-5}M$ 9. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 1.13 \times 10^{- 5})$ $pH = 4.95$ 10. Since we are adding a strong base, this reaction will occur: $CH_3CO_2H(aq) + OH^-(aq) -- \gt CH_3C{O_2}^-(aq) + H_2O(l)$ And these are the concentrations after this reaction: Remember: Reactants at equilibrium = Initial Concentration - y And Products = Initial Concentration + y Since $NaOH$ is a strong base, y = $[NaOH] = 0.021M$ $[CH_3CO_2H] = 0.15 M - 0.021 = 0.13M$ $[CH_3C{O_2}^-] = 0.24M + 0.021 = 0.26M$ 11. Now, calculate the hydronium ion concentration after the addition of the $NaOH$: $[H_3O^+] = Ka * (\frac{[CH_3CO_2H]}{[CH_3C{O_2}^-]})$ $[H_3O^+] = 1.8 \times 10^{-5} * \frac{0.13}{0.26}$ $[H_3O^+] = 1.8 \times 10^{-5} * 0.49$ $[H_3O^+] = 8.9 \times 10^{-6}$ 12. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 8.9 \times 10^{- 6})$ $pH = 5.05$