Question

A buffer is made up of \(239 \mathrm{~mL}\) of \(0.187 \mathrm{M}\) potassium hydrogen tartrate \(\left(\mathrm{KHC}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right)\) and \(137 \mathrm{~mL}\) of \(0.288 \mathrm{M}\) potassium tartrate \(\left(\mathrm{K}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right) . K_{\mathrm{a}}\) for \(\left(\mathrm{H}_{2} \mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{6}\right)\) is \(4.55 \times 10^{-5} .\) Assuming volumes are additive, calculate (a) the \(\mathrm{pH}\) of the buffer. (b) the \(\mathrm{pH}\) of the buffer after adding \(0.0250 \mathrm{~mol}\) of \(\mathrm{HCl}\) to \(0.376 \mathrm{~L}\) of the buffer. (c) the \(\mathrm{pH}\) of the buffer after adding \(0.0250 \mathrm{~mol}\) of \(\mathrm{KOH}\) to \(0.376 \mathrm{~L}\) of the buffer.

Short answer

Question: Calculate the pH of the buffer in various situations: (a) the initial pH of the buffer, (b) the pH after adding a certain amount of HCl, and (c) the pH after adding a certain amount of KOH. Answer: (a) The initial pH of the buffer is 4.23. (b) The pH of the buffer after adding HCl is 2.28. (c) The pH of the buffer after adding KOH is 5.63.

Step-by-step solution

Calculate initial buffer concentrations

Since volumes are additive, we can determine the total volume of the buffer and use the initial volumes and concentrations to calculate the number of moles of each salt in the buffer: Total volume of the buffer solution = 239 mL + 137 mL = 376 mL = 0.376 L Moles of potassium hydrogen tartrate (KHC4H4O6): n = (volume)(concentration) = (0.239 L)(0.187 M) = 0.04469 mol Moles of potassium tartrate (K2C4H4O6): n = (volume)(concentration) = (0.137 L)(0.288 M) = 0.03942 mol Now, calculate their concentrations in the final buffer solution: Concentration of KHC₄H₄O₆ = (0.04469 mol) / (0.376 L) ≈ 0.1187 M Concentration of K₂C₄H₄O₆ = (0.03942 mol) / (0.376 L) ≈ 0.1049 M

Determine the initial pH of the buffer

Now we use the Henderson-Hasselbalch equation to calculate the pH of the buffer: pH = pKa+ log ([A⁻]/[HA]) Here, [HA] is the concentration of potassium hydrogen tartrate and [A⁻] is the concentration of potassium tartrate. The pKa value can be computed as pKa = -log(Ka) = -log(4.55 x 10^(-5)) ≈ 4.34 pH = 4.34 + log(0.1049/0.1187) ≈ 4.34 - 0.11 = 4.23 The initial pH of the buffer is 4.23.

Calculate the pH after adding HCl

We know that 0.0250 mol of HCl is added to the buffer, which will react with potassium tartrate: HCl + K₂C₄H₄O₆ → 2 KCl + H₂C₄H₄O₆ Moles of K₂C₄H₄O₆ consumed = 0.0250 mol (since there are two moles of HCl reacting for every one mole of K₂C₄H₄O₆) Moles of K₂C₄H₄O₆ left = 0.03942 - 0.02500 = 0.01442 mol Moles of H₂C₄H₄O₆ produced = 2 × 0.0250 mol = 0.0500 mol (since there are two moles of HCl reacting for every one mole of K₂C₄H₄O₆) Recalculate the concentrations in the buffer after the reaction with HCl has occurred: [HA] = (0.0500 + 0.04469) / 0.376 ≈ 0.2518 M [A⁻] = (0.01442) / 0.376 ≈ 0.0384 M pH = 4.34 + log(0.0384/0.2518) ≈ 4.34 - 2.06 = 2.28 The pH of the buffer after adding HCl is 2.28.

Calculate the pH after adding KOH

Since 0.0250 mol of KOH is added to the buffer, it will react with potassium hydrogen tartrate: KOH + KHC₄H₄O₆ → K₂C₄H₄O₆ + H₂O Moles of KHC₄H₄O₆ consumed = 0.0250 mol Moles of K₂C₄H₄O₆ produced = 0.0250 mol Recalculate the concentrations in the buffer after the reaction with KOH has occurred: [HA] = (0.04469 - 0.0250)/0.376 ≈ 0.0523 M [A⁻] = (0.0250 + 0.03942) / 0.376 ≈ 0.1711 M pH = 4.34 + log(0.1711/0.0523) ≈ 4.34 + 1.29 = 5.63 The pH of the buffer after adding KOH is 5.63. Summary: (a) The initial pH of the buffer is 4.23. (b) The pH of the buffer after adding HCl is 2.28. (c) The pH of the buffer after adding KOH is 5.63.