Calculate the pH of a solution prepared by mixing 15.0mL of 0.10 M NaOH and 30.0

Question

Calculate the pH of a solution prepared by mixing 15.0mL of 0.10 M NaOH and 30.0mL of 10.0M benzoic acid solution. (benzoic acid is monoprotic; its dissociation constant is 6.3*10 to negative -5).

Explanation / Answer

Start with the chem equation for the reaction: benzoic acid (aq) + OH- (aq) --> benzoate ion (aq) + H2O (l) The benzoate ion is the conjugate base of benzoic acid. Then find the moles of NaOH and benzoic acid: Moles of NaOH = (0.015 L)(0.10 M) = 0.0015 mol Moles of BA = (0.030 L)(0.10 M) = 0.030 mol These two react with a 1:1 stoichiometry, so once the reaction is done, you are left with 0.0015 mol of unreacted benzoic acid and 0.0015 mol of benzoate ion. Then find the concentration of benzoic acid and benzoate ion. For that, you need to total volume, which is now 15.0 mL + 30.0 mL = 45.0 mL, or 0.0450L. Concentrations are: [benzoic acid] = 0.0015 mol / 0.0450L = 0.0333 M [benzoate ion] = 0.0015 mol / 0.0450L = 0.0333 M You have equal concentrations of an acid and its conjugate base, which makes the solution a buffer. You can find the pH from the Henderson Hasselbach equation. pH = pKa + log{[benzoic acid][benzoate ion]} pH = pKa = -log 6.5^10-5 = 4.19. So you were absolutely right.

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