Question

25.0 ml of hydrochloric acid (4.00M) was added to 550.0 ml of a 0.500 M sodium lactate solution. What is the pH of the resulting solution? The pKa for lactic acid is 3.86.

Answer 1

millimoles of sodium lactate = 550 x 0.5 = 275

millimoles of HCl added = 25 x 4.0 = 100

sodium lactate + HCl -----------> lacticacid + sodium chloride

after HCl added

millimoles of sodium lactate = 275 - 100 = 175

millimoles of lactic acid = 100

total volume = 550 + 25 = 575 mL

[lactic acid] = 100 / 575 = 0.174 M

[sodium lactate] = 175 / 575 = 0.304 M

mixture of lactic acid and sodium lactate act as acidic buffer

pH = pKa + log [sodium lactate] / [lactic acid]

pH = 3.86 + log [0.304] / [0.174]

pH = 4.10