Question

What is the pH of a 125mM solution of sodium lactate?

Calculate the pH of a 0.579 M aqueous solution of Lactic Acid (pKa = 3.76) Which of the following acids, if in solutions of equal concentration, is the least acidic? a) 0.1M propanoic acid pKa = 4.8 b) 0.1M octanoic acid pKa = 4.89 c) 0.1 M uric acid, pKa = 3.89 d) 0.1 M formic acid pKa = 3.75 e) All of these acids are equally acidic because they are all of equal concentration. Which of the following acids is the least acidic? a) 0.1M propanoic acid pka = 4.86 b) 0.5M octanoic acid pKa = 4.89 c)0.01M uric acid, pKa = 3.89 d) 0.03 M formic acid pKa = 3.75 e) All of these acids are equally acidic because they are all weak acids If 30ml of 100 mM NaOH is added to 100ml of 150mM Acetic Acid (pKa = 4.75), what is the final pH? What is the pH of a 125 mM solution of Sodium Lactate? (Use pKa above)

Answer 1

What is the pH of a 125mM solution of sodium lactate?

Let the lactic acid be HA

use:

pKa = -log Ka

3.76 = -log Ka

Ka = 1.738*10^-4

find kb for lactate ion (A-)

use:

Kb = (1.0*10^-14)/Ka

Kb = (1.0*10^-14)/1.738*10^-4

Kb = 5.754*10^-11

A- dissociates as

A- + H2O -----> HA + OH-

0.125 0 0

0.125-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((5.754*10^-11)*0.125) = 2.682*10^-6

since c is much greater than x, our assumption is correct

so, x = 2.682*10^-6 M

use:

pOH = -log [OH-]

= -log (2.682*10^-6)

= 5.57

use:

PH = 14 - pOH

= 14 - 5.57

= 8.43

Answer: 8.43