Question

In: Chemistry

calculate the pH of the following solutions. a) 0.20 M Na2HPO4 b) 0.05 M NaHSO4 c)...

calculate the pH of the following solutions.

a) 0.20 M Na2HPO4

b) 0.05 M NaHSO4

c) 4.567 grams of NaH2PO4 dissolved in 1.50 liters of solution

Expert Solution

a]

Na2HPO4 ----> 2Na+ + HPO4-2

HPO4-2 ----------> H+ + PO4-3

0.2 0 0

0.2-x x x

Ka3 of H3PO4 = 4.5*10^-13

Ka3 = [H+] [PO4-3] / [HPO4-2]

4.5*10^-13 = x^2 / (0.2-x)

x^2 + 4.5*10^-13 x - 9*10^-14 = 0

x = 3*10^-7

[H+] = 3*10^-7

pH = -log [H+] = 6.522

b]

NaHSO4 ---> Na+ + HSO4-

HSO4- -----> H+ + SO4-2

0.05 0 0

0.05-x x x

Ka2 = x^2 / (0.05-x) = 1.2*10^-2

x^2 + 1.2*10^-2 x - 6*10^-4 = 0

x = 0.0192

pH = -log x = 1.7166

c]

Molarity = mass / Molarmass*V in L

Molar mass of NaH2PO4 = 120 gms

Molarity = 4.567 / (120*1.5) = 0.02537 M

H2PO- ----------> H+ + HPO4-2

0.02537 0 0

0.02537-x x x

Ka2 = x^2 / (0.02537-x) = 6.3*10^-8

x^2 + 6.3*10^-8 x - 1.598*10^-9 = 0

x = 3.994*10^-5

pH = -log x = 4.398

queen_honey_blossom answered 2 years ago

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