Question

a) What is the pH of 0.10 M NaH₂PO₄? K_a2 = 6.2 x 10^-8

b) What is the pH of 0.10 M Na₂​HPO₄? K_a3 = 2.2 x 10^-13

Answer 1

a)

H2PO4- dissociates as:

H2PO4- -----> H+ + HPO42-

0.1 0 0

0.1-x x x

Ka = [H+][HPO42-]/[H2PO4-]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.2*10^-8)*0.1) = 7.874*10^-5

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

so, x = 7.874*10^-5 M

So, [H+] = x = 7.874*10^-5 M

use:

pH = -log [H+]

= -log (7.874*10^-5)

= 4.1038

Answer: 4.10

b)

HPO42- dissociates as:

HPO42- -----> H+ + PO43-

0.1 0 0

0.1-x x x

Ka = [H+][PO43-]/[HPO42-]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((2.2*10^-13)*0.1) = 1.483*10^-7

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

so, x = 1.483*10^-7 M

So, [H+] = x = 1.483*10^-7 M

use:

pH = -log [H+]

= -log (1.483*10^-7)

= 6.8288

Answer: 6.83