Acid/Base Titration

(Titration of a base with an acid)

Problem:

Calculate the molarity of an acetic acid solution if 34.57 mL of this solution are needed to neutralize 25.19 mL of 0.1025 M sodium hydroxide

CH3COOH (aq) + NaOH (aq) -----> Na+(aq) + CH3COOH-(aq) + H2O (l)

Strategy:

  1. Figure out how many moles of the titrant (in this case, the base) were needed.

    (25.19 mL) (1 L/ 1000 mL) = 0.02519 L

    (0.02519 L) (0.1025 mol / L) = 0.002582 mol NaOH

  2. Use the balanced chemical equation to calculate the moles of analyte (in this case, the acid) present.

    (0.002582 mol NaOH) (1 mol CH3COOH / 1 mol NaOH) = 0.002582 mol CH3COOH

  3. Use the volume of analyte to find the concentration of the analyte.

    (34.57 mL) (1L / 1000 mL) = 0.03457 L

    (0.002582 mol CH3COOH / 0.03457 L) = 0.07469 M


Previous Step

Stoichiometry Problems Problem Solving General Chemistry Help Homepage