# Sample problems buffer solution

1. Problem: A buffer solution is prepared by mixing 50 mL of 0.1 M CH₃COOH (acetic acid, pKa = 4.74) with 50 mL of 0.1 M CH₃COONa (sodium acetate). Calculate the pH of the buffer. Solution: Use the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). [A⁻] = [CH₃COO⁻] = 0.1 M, [HA] = [CH₃COOH] = 0.1 M. Substituting the values gives pH = 4.74 + log(1) = 4.74.

2. Problem: A buffer composed of equal concentrations of NH₃ (ammonia, pKb = 4.75) and NH₄Cl (ammonium chloride) has a pH of 9.25. What is the pH after addition of 0.01 mol of HCl to 1 L of the buffer? Solution: Since HCl is a strong acid, it will react with NH₃ to form NH₄⁺, decreasing the concentration of NH₃ and increasing the concentration of NH₄⁺. The new pH can be calculated using the Henderson-Hasselbalch equation: pH = pKw – pKb + log([NH₃]/[NH₄⁺]) = 14 – 4.75 + log((0.1-0.01)/(0.1+0.01)) = 9.16.
3. Problem: Calculate the pH of a buffer solution containing 0.1 M benzoic acid (pKa = 4.20) and 0.15 M sodium benzoate. Solution: Apply the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]) = 4.20 + log(0.15/0.1) = 4.38.
4. Problem: A buffer solution contains 0.2 M acetic acid and 0.3 M sodium acetate. What is the pH after adding 0.01 mol of NaOH to 1 L of this buffer? Solution: NaOH will react with acetic acid to form acetate, increasing the concentration of acetate and decreasing the concentration of acetic acid. New pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = 4.74 + log((0.3+0.01)/(0.2-0.01)) = 4.87.
5. Problem: A buffer solution is prepared by mixing 100 mL of 0.2 M HCN (pKa = 9.31) and 200 mL of 0.1 M NaCN. What is the pH of the buffer? Solution: The concentrations of HCN and CN⁻ are [HCN] = (0.2 x 100) / (100+200) = 0.067 M, [CN⁻] = (0.1 x 200) / (100+200) = 0.067 M. The pH = pKa + log([CN⁻]/[HCN]) = 9.31 + log(1) = 9.31.
6. Problem: Calculate the pH of a buffer solution made by mixing 50 mL of 0.1 M HCl with 50 mL of 0.2 M sodium acetate. Solution: HCl will react with acetate to form acetic acid. The new concentration of acetic acid = 0.05 mol/L, and [acetate] = 0.05 mol/L. The pH = pKa + log([acetate]/[acetic acid]) = 4.74 + log(1) = 4.74.
7. Problem: Determine the pH of a buffer made by adding 0.01 mol of solid NH₄Cl to 1 L of a solution of 0.1 M NH₃. Solution: NH₄Cl will dissolve and increase the [NH₄⁺]. The new pH = pKw – pKb + log([NH₃]/[NH₄⁺]) = 14 – 4.75 + log((0.1)/(0.01)) = 9.50.
8. Problem: What is the pH of a solution made by mixing 100 mL of 0.2 M NH₄Cl and 100 mL of 0.1 M NH₃? Solution: [NH₄⁺] = (0.2 x 100) / (100+100) = 0.1 M, [NH₃] = (0.1 x 100) / (100+100) = 0.05 M. The pH = pKw – pKb + log([NH₃]/[NH₄⁺]) = 14 – 4.75 + log(0.05/0.1) = 8.75.
9. Problem: A buffer solution contains 0.25 M acetic acid and 0.3 M sodium acetate. What is the pH after adding 0.02 mol of HCl to 1 L of this buffer? Solution: HCl will react with acetate to form acetic acid, increasing the concentration of acetic acid and decreasing the concentration of acetate. New pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = 4.74 + log((0.3-0.02)/(0.25+0.02)) = 4.68.
10. Problem: Determine the pH of a buffer solution made by adding 0.01 mol of solid CH₃COONa to 1 L of a solution of 0.1 M CH₃COOH. Solution: CH₃COONa will dissolve and increase the [CH₃COO⁻]. The new pH = pKa + log([CH₃COO⁻]/[CH₃COOH]) = 4.74 + log((0.01)/(0.1)) = 4.24.
11. Problem: Calculate the pH of a buffer solution prepared by mixing 100 mL of 0.1 M HNO₂ (pKa = 3.37) with 100 mL of 0.2 M NaNO₂. Solution: Use the Henderson-Hasselbalch equation: pH = pKa + log([NO₂⁻]/[HNO₂]) = 3.37 + log(0.2/0.1) = 3.67.
12. Problem: A buffer solution contains 0.1 M HF (pKa = 3.17) and 0.2 M NaF. What is the pH of the solution? Solution: Using the Henderson-Hasselbalch equation: pH = pKa + log([F⁻]/[HF]) = 3.17 + log(0.2/0.1) = 3.47.
13. Problem: Calculate the pH of a buffer solution containing 0.2 M HCN and 0.1 M NaCN. Solution: Use the Henderson-Hasselbalch equation: pH = pKa + log([CN⁻]/[HCN]) = 9.31 + log(0.1/0.2) = 8.81.
14. Problem: How much solid NH₄Cl (in moles) should be added to 1 L of 0.1 M NH₃ solution to get a buffer of pH 9.0? Solution: Use the Henderson-Hasselbalch equation: pH = pKw – pKb + log([NH₃]/[NH₄⁺]). Solve it for [NH₄⁺]: [NH₄⁺] = [NH₃] x 10^(pKw – pKb – pH) = 0.1 x 10^(14 – 4.75 – 9) = 0.0056 M. So, 0.0056 moles of NH₄Cl should be added.
15. Problem: How much solid CH₃COONa (in moles) should be added to 1 L of 0.1 M CH₃COOH solution to make a buffer of pH 5.0? Solution: Use the Henderson-Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). Solve it for [A⁻]: [A⁻] = [HA] x 10^(pH – pKa) = 0.1 x 10^(5 – 4.74) = 0.056 M. So, 0.056 moles of CH₃COONa should be added.
16. Problem: Determine the pH of a buffer solution containing 0.2 M NH₄Cl and 0.1 M NH₃. Solution: Use the Henderson-Hasselbalch equation: pH = pKw – pKb + log([NH₃]/[NH₄⁺]) = 14 – 4.75 + log(0.1/0.2) = 8.75.
17. Problem: Calculate the pH of a buffer solution that contains 0.15 M HC₂H₃O₂ and 0.1 M NaC₂H₃O₂. Solution: Use the Henderson-Hasselbalch equation: pH = pKa + log([C₂H₃O₂⁻]/[HC₂H₃O₂]) = 4.74 + log(0.1/0.15) = 4.52.
18. Problem: A buffer solution is prepared by mixing 200 mL of 0.2 M HNO₂ (pKa = 3.37) and 200 mL of 0.1 M NaNO₂. What is the pH of the buffer? Solution: [HNO₂] = (0.2 x 200) / (200+200) = 0.1 M, [NO₂⁻] = (0.1 x 200) / (200+200) = 0.05 M. pH = pKa + log([NO₂⁻]/[HNO₂]) = 3.37 + log(0.05/0.1) = 3.07.
19. Problem: How much 0.1 M NaOH (in liters) should be added to 1 L of a buffer containing 0.1 M acetic acid and 0.1 M sodium acetate to achieve a pH of 5.0? Solution: The new [acetate] and [acetic acid] can be calculated from the Henderson-Hasselbalch equation: pH = pKa + log([acetate]/[acetic acid]). Solving it for [acetate] gives [acetate] = [acetic acid] x 10^(pH – pKa) = 0.1 x 10^(5 – 4.74) = 0.176 M. The volume of NaOH needed = (0.176 – 0.1) M / 0.1 M/L = 0.76 L.
20. Problem: Calculate the pH of a buffer solution prepared by mixing 100 mL of 0.1 M HCl and 100 mL of 0.2 M NH₄OH. Solution: HCl will react with NH₄OH to form NH₄Cl, decreasing [NH₄OH] and increasing [NH₄⁺]. [NH₄⁺] = (0.1 x 100) / (100+100) = 0.05 M, [NH₄OH] = (0.2 x 100) / (100+100) = 0.1 M. The pH = pKw – pKb + log([NH₄OH]/[NH₄⁺]) = 14 – 4.75 + log(0.1/0.05) = 9.25.