Question

Lactic acid, CH3CH(OH)COOH, is found in sour milk. A solution containing 1.53 g NaCH3CH(OH)COO in 200.0 mL of 0.0500 M CH3CH(OH)COOH has a pH= 4.00. What is Ka of lactic acid? Express your answer using two significant figures.

Answer 1

[CH3CH(OH)COOH] = 0.0500 M

Molar mass of NaC3H5O3,

MM = 1*MM(Na) + 3*MM(C) + 5*MM(H) + 3*MM(O)

= 1*22.99 + 3*12.01 + 5*1.008 + 3*16.0

= 112.06 g/mol

mass(NaC3H5O3)= 1.53 g

use:

number of mol of NaC3H5O3,

n = mass of NaC3H5O3/molar mass of NaC3H5O3

=(1.53 g)/(1.121*10^2 g/mol)

= 1.365*10^-2 mol

volume , V = 2*10^2 mL

= 0.2 L

use:

Molarity,

M = number of mol / volume in L

= 1.365*10^-2/0.2

= 6.827*10^-2 M

so,

[NaCH3CH(OH)COO] = 6.827*10^-2 M

use:

pH = pKa + log ([NaCH3CH(OH)COO] / [CH3CH(OH)COOH])

4.00 = pKa + log ((6.827*10^-2)/0.0500)

4.00 = pKa + log (1.3654)

pKa = 3.865

use:

pKa = -log Ka

3.865 = -log Ka

Ka = 1.365*10^-4

Answer: 1.4*10^-4