Question

A cheese maker added 4.00 L of milk to a sample of bacterial culture. He knew that once the concentration of lactic acid reached 1.25 x 10^-2 mol/L the cheese would be ready. He took a 10.00 mL sample of the cheese, added several drops of phenolphthalein, and titrated the sample against a 0.111 mol/L NaOH solution. 1.15 mL of NaOH was required to for equivalence.

CH3CHOHCOOH(aq) + NaOH(aq) --> H2O(l) + CH3CHOHCOONa(aq)

Determine the concentration of lactic acid in the cheese.

Answer 1

V(cheese) = 10 mL

C(NaOH) = 0.111 mol/L

V(NaOH) = 1.15 mL

Clactic acid = C(NaOH)*V(NaOH)/V(cheese) = 0.111mol/L*1.15mL/10mL = 0.012765 mol/L = 1.2765·10-2 mol/L