Question

At 25 ∘C phosphoric acid, H3PO4, has the following equilibrium constants:

H3PO4(aq)+H2O(l)H2PO4−(aq)+H2O(l)HPO42−(aq)+H2O(l)⇌⇌⇌H3O+(aq)+H2PO−4(aq)H3O+(aq)+HPO42−(aq)H3O+(aq)+PO43−(aq

What is the pH of a solution of 0.300 M K2HPO4, potassium hydrogen phosphate? Express your answer numerically to the hundredths place.

What is the pH of a solution of 0.700 M KH2PO4, potassium dihydrogen phosphate?

Express your answer numerically to the hundredths place.

Answer 1

The given salt K2HPO4 is a strong salt and dissociates as follows

K2HPO4 -------> 2 K⁺ + HPO₄^-2

HPO4^2- can accept or donate a proton. The reaction that favors depends on equilibrium constant Ka or Kb

HPO4^2- + H2O <------------------> H3O+ + PO4 ^3- Ka3 = 4.2 x 10^-13

HPO4^2- + H2O <--------------------> OH- + H2PO4^- (Kw/Ka2) = Kb = 1.61 x 10^-7

Since Kb value is higher than Ka, base dissociation favors.

Let's draw ICE table for the given reaction

HPO4^2- + H2O <--------------------> OH- + H2PO4^-

	HPO4^2-	OH-	H2PO4^-
I	0.300	0	0
C	-x	+x	+x
E	0.300 - x	x	x

The equilibrium constant can be written as

Kb = [OH-] [H2PO4^-] / [HPO4^2-]

1.61 x 10^-7 = x^2 / 0.300 - x ( we can use approximation as value of Kb is very small)

1.61 x 10^-7 = x^2 / 0.300

x^2 = 4.83 x 10^-8

x = 2.198 x 10^-4

[OH-] = 2.198 x 10^-4 M

pOH = - log [OH]

pOH = - log ( 2.198 x 10^-4)

pOH = 3.66

pH = 14 - 3.66

pH = 10.34

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The given salt dissociates as follows

KH2PO4 -------------> K+ + H2PO4^-

H2PO4^- can accept or donate a proton. The reaction that favors depends on Ka/Kb value.

H2PO4^- + H2O <-------------------> H3O+ + HPO4^2- Ka2 = 6.2×10−8

H2PO4^- + H2O <-------------------> OH- + H3PO4 Kb = kw/Ka1 = 1.33 x 10^-12

Since Ka is higher than Kb, acid dissociation favors.

The ICE table can be drawn as follows

H2PO4^- + H2O <-------------------> H3O+ + HPO4^2-

	H2PO4^-	H3O+	HPO4^2-
I	0.700	0	0
C	-x	+x	+x
E	0.700 - x	x	x

Equilibrium constant can be written as

Ka2 = [H3O+] [ HPO4^2-] / [H2PO4^-]

6.2 x 10^-8 = x^2 / 0.700 - x .........( we can use approximation )

6.2 x 10^-8 = x^2/0.700

x^2 = 4.34 x 10^-8

x = 2.08 x 10^-4

[H3O+] =  2.08 x 10^-4

pH = - log [H3O+]

pH = - log ( 2.08 x 10^-4)

pH = 3.68