In: Chemistry
Lactic acid, C3H6O3, occurs in sour milk as a result of the metabolism of certain bacteria. Calculate the pH of a solution of 56 mg lactic acid in 250.0 mL water. Ka for D-lactic acid is 1.4 × 10-4.
I mostly need help with solving the quadratic formula portion of this problem
C3H6O3 <-----> C3H5O3- + H+
molarity of lactic acid = (56*10^-3)/90 *1000/250
= 0.0025 M
C3H6O3 <-----> C3H5O3- + H+
0.0025 0 0 initial
-x +x +x change
0.0025-x x x equilibrium
Ka = x^2/(0.0025-x)
1.4*10^-4 = x^2/(0.0025-x)
x = 0.000526
pH = -log(0.000526)
= 3.28