## Chemistry and Chemical Reactivity (9th Edition)

(a) $pH (Buffer) = 4.54$ (b) The pH of the buffer is higher. (Because we added a base.) $pH (Lactic acid)= 2.43$
1. Calculate the molar mass: 22.99* 1 + 12.01* 1 + 1.01* 3 + 12.01* 1 + 1.01* 1 + 16* 1 + 1.01* 1 + 12.01* 1 + 16* 2 = 112.07g/mol 2. Calculate the number of moles $n(moles) = \frac{mass(g)}{mm(g/mol)}$ $n(moles) = \frac{2.75}{ 112.07}$ $n(moles) = 0.0245$ 3. Find the concentration in mol/L: $C(mol/L) = \frac{n(moles)}{volume(L)}$ $C(mol/L) = \frac{ 0.0245}{ 0.05}$ $C(mol/L) = 0.491$ 4. Calculate the pKa Value $pKa = -log(Ka)$ $pKa = -log( 1.4 \times 10^{- 4})$ $pKa = 3.85$ 5. Check if the ratio is between 0.1 and 10: - $\frac{[Base]}{[Acid]} = \frac{0.491}{0.1}$ - 4.91: It is. 6. Check if the compounds exceed the $K_a$ by 100 times or more: - $\frac{0.491}{1.4 \times 10^{-4}} = 3505$ - $\frac{0.1}{1.4 \times 10^{-4}} = 714$ 7. Using the Henderson–Hasselbalch equation: $pH = pKa + log(\frac{[Base]}{[Acid]})$ $pH = 3.85 + log(\frac{0.491}{0.1})$ $pH = 3.85 + log(4.91)$ $pH = 3.85 + 0.691$ $pH = 4.54$ ----- Calculating the pH of the pure acid solution: 1. We have those concentrations at equilibrium: -$[H_3O^+] = [CH_3CHOHC{O_2}^-] = x$ -$[CH_3CHOHCO_2H] = [CH_3CHOHCO_2H]_{initial} - x = 0.1 - x$ For approximation, we consider: $[CH_3CHOHCO_2H] = 0.1M$ 2. Now, use the Ka value and equation to find the 'x' value. $Ka = \frac{[H_3O^+][CH_3CHOHC{O_2}^-]}{ [CH_3CHOHCO_2H]}$ $Ka = 1.4 \times 10^{- 4}= \frac{x * x}{ 0.1}$ $Ka = 1.4 \times 10^{- 4}= \frac{x^2}{ 0.1}$ $1.4 \times 10^{- 5} = x^2$ $x = 3.74 \times 10^{- 3}$ Percent dissociation: $\frac{ 3.74 \times 10^{- 3}}{ 0.1} \times 100\% = 3.74\%$ %dissociation < 5% : Right approximation. Therefore: $[H_3O^+] = [CH_3CHOHC{O_2}^-] = x = 3.74 \times 10^{- 3}M$ 3. Calculate the pH Value $pH = -log[H_3O^+]$ $pH = -log( 3.74 \times 10^{- 3})$ $pH = 2.43$