Question

Suppose you disolve 1.2 g of KH2PO4 in 1L of water and then adjust the pH of the solution to 6.8. What are the concentrations of the species listed below?

A) [H3PO4]

B) [H2PO−4]

C) [HPO−24]

D) [PO−34]

Answer 1

pH of the KH₂PO4 solution = 6.8

Moles of KH₂PO4 = 1.2 / 136.085 = 0.0088 moles

Molarity of KH₂PO4 = 0.0088/1 = 0.0088 M

The pH of a buffer system can be calculated using the Henderson-Hasselbalch equation.

pH = pKa + log [A^-] / [HA]

This equation can also be used to calculate amounts of acid and base required to make a buffer

of a certain pH and with an acid whose pKa is known. Accordingly when the above equation is

rearranged one will get:

10^{pH – pKa} = [base] / [Acid]

Thus we need to use three pKa values of phosphoric acid from literature to obtain the molarity

of individual components of the buffer.

Phosphoric acid pKa values from literature are given below:

pKa1 for H₃PO₄ and its conjugate base H₂PO₄^- = 2.12

pKa2 for H₂PO₄^- and its conjugate base HPO₄^2-= 7.21

pKa3 for HPO₄^2-and its conjugate base PO₄^3-= 12.32

For finding molarity of H₃PO₄ one has to use pKa1 (2.12) as given below:

[H₂PO₄^-]/[ H₃PO₄] = 10^6.8-2.12

[H₂PO₄^-]/[ H₃PO₄] = 10^4.68

[H₂PO₄^-]/[ H₃PO₄] = 47683/1

[H₂PO₄^-] = 47683 parts

[H₃PO₄] = 1 part

Whole of H₂PO₄^- and H₃PO₄ = 47683 + 1 = 47684

Then calculate the decimal fraction (part/whole) of each buffer component.

H₂PO₄^- = 47683/47684 = 0.99998

H₃PO₄ = 0.00002

Find the molarity (M) of each component in the buffer by simply multiplying the molarity of  

the buffer by the decimal fraction of each component.

[H₂PO₄^-] = 0.99998 x 0.0088 = 0.008799 = 8.799 x 10^-3 M

[H₃PO₄] = 0.00002 x 0.0088 = 1.76 x 10^-7 M

For finding molarity of H₂PO₄^- andHPO₄^2- one has to use pKa2 (7.21) as given below:

[HPO₄^2-]/[H₂PO₄^-] = 10^6.8-7.21

[HPO₄^2-]/[H₂PO₄^-] = 10^-0.41

[HPO₄^2-]/[H₂PO₄^-] = 2.57/1

Whole of HPO₄^2- andH₂PO₄^- = 2.57+1=3.57

Then calculate the decimal fraction (part/whole) of each buffer component as below.

[H₂PO₄^-] = 1/3.57 = 0.28

[HPO₄^2-] = 2.57/3.57 = 0.72

Find the molarity (M) of each component in the buffer by simply multiplying the molarity of

       the buffer by the decimal fraction of each component.

[H₂PO₄^-] = 0.28 x 0.0088 = 0.002464 = 2.464 x 10^-3 M

[HPO₄^2-] = 0.72 x 0.0088 = 0.006336 = 6.336 x 10^-3 M

For finding molarity of PO₄^3-one has to use pKa3 (12.32) as given below:

[PO₄^3-]/[HPO₄^2-]= 10^6.8-12.32

[PO₄^3-]/[HPO₄^2-]= 10^-5.52

[PO₄^3-]/[HPO₄^2-]= 3.02 x 10^-6/1

Whole of PO₄^3- andHPO₄^2- = 3.02 x 10^-6 and 1.0 = 1.00000302

PO₄^3- = 3.02 x 10^-6/1.00000302 = 0.00000302

[PO₄^3-] = 0.00000302 x 0.0088 = 2.657 x 10^-8 M

HPO₄^2- = 1/1.00000302 = 0.99999698

[HPO₄^2-] = 0.99999698 x 0.0088 = 8.799 x 10^-3 M

Thus the molarities of each component of the KH2PO4 buffer of pH 6.8 are as follows:

H₃PO₄ = 1.76 x 10^-7 M

H₂PO₄^- = 2.464 x 10^-3 M

HPO₄^2- = 6.336 x 10^-3 M

PO₄^3- = 2.657 x 10^-8 M