Answer to Question #209664 in Chemistry for shaza

Question

Lactic acid, C3H6O3, is built in human muscles during anaerobic exercises. If the initial concentration of lactic acid is 0.12 mol/L and the pH is 2.39, what is the Ka for the lactic acid?

Answer 1

The pH is telling us the concentration of the H₃O⁺_(aq) and C₃H₅O₃⁻_(aq)

pH =–log [H₃O_+(aq)]

[H₃O⁺] =10^{– pH} = 10^–2.39 = 4.1 × 10^–3 mol/L

Ka = ([H₃O⁺][C₃H₅O₃⁻]) / [C₃H₆O₃] = ((4.1×10⁻³)(4.1×10⁻³)) / 0.116 = 1.45 × 10⁻⁴

Therefore, Ka is 1.45 × 10⁻⁴.