Question

a. What is the pH of a 0.25 M lactic acid (CH3(CH2)COOH) solution? Known: pKa = 3.85
I set the equation up as CH3(CH2)COOH –> CH3(CH2)COO- + H3O+. My final answer I got was pH = 3.99.

b. Assume you had enough solid sodium lactate (NaCH3CH(OH)COO) to cause the lactate ion (CH3CH(OH)COO-) concentration in the solution to increase to 0.20 M. what is the pH of the solution after the addition of sodium lactate to lactic acid?
I have no idea what to do with this one, but I’m assuming Na is a spectator ion?

Please show the work necessary for part b, and part a if my answer is incorrect. Thanks.

Answer 1

a)

CH3CH(OH)COOH + H2O   ------------------> CH3CH(OH)COO ^- + H3O⁺

        0.25                                                                 0                         0        -------- at initial

    0.25 - x                                                                x                          x      ---------- at equilibrium

pKa = 3.85

ka = 10^-pKa

    = 10^-3.85

    = 1.41 x 10^-4

ka = [CH3CH(OH)COO ^-][H3O⁺] / [CH3CH(OH)COOH]

1.41 x 10^-4 = x^2 / 0.25 - x

x = 5.87 x 10^-3

[H3O⁺] = 5.87 x 10^-3 M

pH = -log [H3O+]

pH = -log (5.87 x 10^-3)

pH = 2.23

B)

acid = 0.25 M

salt sodium lactate = 0.2 M

lactic acid + sodium lactate mixture can act as a buffer. so we have to use buffer pH formula

pH = pKa + log [salt / acid ]

     = 3.85 + log [0.2/0.25]

pH     = 3.75