What is the pH of a solution of 0.15 M formic acid?

What is the pH of the solution that results from adding 30.0 mL of 0.015 M KOH to 50.0 mL of 0.015 M C7H6O2?

• 30.0 mL of 0.015 M KOH 50.0 mL of 0.015 M benzoic acid please solve

• Answer:

  This is a buffer problem. The buffer has a weak acid, C6H5COOH, and its conjugate base, C6H5COO^-. moles C6H5COOH = (0.0500 L)(0.015 M) = 0.00075 mole moles KOH = V*M = (0.0300 L)(0.015 M) = 0.00045 mole Reaction: C6H5COOH(aq) + OH^-(aq) --> C6H5COO^-(aq) + H2O(l) moles C6H5COOH after reaction = 0.00075 mole - 0.00045 mole = 0.00030 mole moles C6H5COO^- after reaction = 0.00045 mole Calculate new concentrations. Total volume of solution = 80.0 mL = 0.0800 L molarity = (moles solute)/(L of solution) [C6H5COOH] = (0.00030 mole)/(0.0800 L) = 0.00375 mol/L [C6H5COO^-] = (0.00045 mole)/(0.0800 L) = 0.005625 mol/L Equilibrium: C6H5COOH(aq) + H2O(l) <=> H3O^+(aq) + C6H5COO^-(aq) Equilibrium constant expression: Ka = [H3O^+][C6H5COO^-]/[C6H5COOH] = 6.28 × 10^–5 From the equilibrium reaction you see that one H3O^+ and one C6H5COO^- are produced each time benzoic acid reacts with water. The best (and easiest) way to calculate the pH is to use the Henderson-Hasselbalch equation. pH = pKa + log{[C6H5COO^-]/[C6H5COOH]} pKa = -logKa = -log(6.28 × 10^–5) = 4.202 pH = 4.202 + log(0.005625/0.00375) = 4.202 + 0.176 = 4.378 = 4.38 Hope this is helpful to you. JIL HIR