A lactic acid solution is prepared by dissolving 0.67 g of lactic acid (HC3H5O3) in 100.00 mL of water. The pH is found to be 2.50. Calculate the Ka for lactic acid.
Molar mass of C3H6O3,
MM = 3*MM(C) + 6*MM(H) + 3*MM(O)
= 3*12.01 + 6*1.008 + 3*16.0
= 90.078 g/mol
mass(C3H6O3)= 0.67 g
use:
number of mol of C3H6O3,
n = mass of C3H6O3/molar mass of C3H6O3
=(0.67 g)/(90.08 g/mol)
= 7.438*10^-3 mol
volume , V = 1*10^2 mL
= 0.1 L
use:
Molarity,
M = number of mol / volume in L
= 7.438*10^-3/0.1
= 7.438*10^-2 M
use:
pH = -log [H+]
2.5 = -log [H+]
[H+] = 3.162*10^-3 M
HC3H5O3 dissociates as:
HC3H5O3 -----> H+ + C3H5O3-
7.438*10^-2 0 0
7.438*10^-2-x x x
Ka = [H+][C3H5O3-]/[HC3H5O3]
Ka = x*x/(c-x)
Ka = 3.162*10^-3*3.162*10^-3/(0.07438-3.162*10^-3)
Ka = 1.404*10^-4
Answer: 1.40*10^-4
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