Question

Suppose you disolve 1.2 g of KH2PO4 in 1L of water and then adjust the pH...

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) [H3PO4^-]

C) [HPO4^−2]

D) [PO4^−3]

PLEASE DO NOT ANSWER if you are not sure!!!

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Homework Answers

Answer #1

mol of KH2PO4 = mass/MW = 1.2/136.085541 = 0.0088179 mol

[H2PO4-] = 0.0088179 M

pH = 6.8

then

pH= pKa1 + log(H2PO4- / H3PO4)

pH= pKa2 + log(HPO4-2 / H2PO4-)

pH= pKa3 + log(PO4-3 / HPO4-2)

pKa data --> 2.12, 7.21, 12.3

substitute known data

[H2PO4-] = 0.0088179

6.8 = 2.12+ log(0.0088179 / H3PO4)

6.8 = 7.21+ log(HPO4-2 / 0.0088179 )

6.8 = 12.3 + log(PO4-3 / HPO4-2)

solve for all:

6.8 = 2.12+ log(0.0088179 / H3PO4)

0.0088179 / (10^(6.8-2.12)) = H3PO4 = 1.8423*10^-7 M

[HPO4-2] = 0.0088179

6.8 = 7.21+ log(HPO4-2 / 0.0088179 )

HPO4-2 = 0.0088179 *10^(6.8 -7.21)

HPO4-2 = 0.003430 M

finally

6.8 = 12.3 + log(PO4-3 / 0.003430)

[PO4-3] = 0.003430*10^(6.8-12.3)

[PO4-3] = 1.084*10^-8

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