a. What is the pH of a 0.25 M lactic acid (CH3(CH2)COOH) solution? Known: pKa =
ID: 880609 • Letter: A
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.
Explanation / Answer
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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.