Question

A buffer contains 0.15 mol of propionic acid $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)$ and 0.10 mol of sodium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\right)\) in 1.20 \(\mathrm{L}\) . (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addition of 0.01 \(\mathrm{mol}\) of \(\mathrm{NaOH}\) ? (c) What is the pH of the buffer after the addition of 0.01 \(\mathrm{mol}\) of \(\mathrm{HI} ?\)

Short answer

(a) The initial pH of the buffer is \(4.87 + log\frac{0.10/1.20}{0.15/1.20}\). (b) After the addition of 0.01 mol of NaOH, the pH of the buffer is \(4.87 + log\frac{0.11/1.20}{0.14/1.20}\). (c) After the addition of 0.01 mol of HI, the pH of the buffer is \(4.87 + log\frac{0.09/1.20}{0.16/1.20}\).

Step-by-step solution

Calculate the initial pH of the buffer

Using the initial amounts of propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\) and sodium propionate \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COONa}\right)\), we will apply the Henderson-Hasselbalch equation to find the pH of the buffer. The equation is: \(pH = pK_a + log\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\). Given: - Initial moles of \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH}\ (HA)\) = 0.15 mol - Initial moles of \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO^-}\ (A^-)\) (from sodium propionate) = 0.10 mol - Volume of buffer = 1.20 L - pKa of propionic acid = 4.87 Calculate concentrations of HA and A^-: - Initial conc. of \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH}\ (HA)\) = \(\frac{0.15}{1.20} \) M - Initial conc. of \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO^-}\ (A^-)\) = \(\frac{0.10}{1.20} \) M Finally, plug the values into the Henderson-Hasselbalch equation: \(pH = 4.87 + log\frac{0.10/1.20}{0.15/1.20}\)

Calculate the pH of the buffer after the addition of 0.01 mol of NaOH

Adding 0.01 mol of NaOH, a strong base, will react with the weak acid in the buffer: \(\mathrm{OH^{-}} + \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH} \rightarrow \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO^{-}} + \mathrm{H}_{2}\mathrm{O}\) The reaction will consume 0.01 mol of HA and produce 0.01 mol of A^-. Update the amounts of HA and A^-: - Moles of HA = 0.15 - 0.01 = 0.14 mol - Moles of A^- = 0.10 + 0.01 = 0.11 mol Now, apply the Henderson-Hasselbalch equation again to calculate the new pH: \(pH = 4.87 + log\frac{0.11/1.20}{0.14/1.20}\)

Calculate the pH of the buffer after the addition of 0.01 mol of HI

Adding 0.01 mol of HI, a strong acid, will react with the conjugate base in the buffer: \(\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COO^-} + \mathrm{H^{+}} \rightarrow \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{COOH}\) The reaction will consume 0.01 mol of A^- and produce 0.01 mol of HA. Update the amounts of HA and A^-: - Moles of HA = 0.15 + 0.01 = 0.16 mol - Moles of A^- = 0.10 - 0.01 = 0.09 mol Finally, apply the Henderson-Hasselbalch equation again to calculate the new pH: \(pH = 4.87 + log\frac{0.09/1.20}{0.16/1.20}\)