Question

Given the following Ka values: H3PO4 <--> H2PO4- + H+, Ka = 7.11*10-3 H2PO4- <--> HPO42-...

Given the following Ka values:

H3PO4 <--> H2PO4- + H+, Ka = 7.11*10-3

H2PO4- <--> HPO42- + H+, Ka = 6.34*10-8

HPO42- <--> PO43- + H+, Ka = 4.22*10-13,

a) please calculate the pH of the solution at the first equivalence point for the above titration.

b) In the same titration as described above, what's the pH of the solution after 15.00 mL of NaOH was added?

Homework Answers

Answer #1

a)

at first equivelence point

pKa1 = -log (7.11 x 10^-3) = 2.15

pKa2 = -log (6.34 x 10^-8) = 7.20

pH = (pKa1 + pKa2 ) / 2

pH = (2.15 + 7.20) / 2

pH = 4.68

b)

millimoles of H3PO4 = M x V

millimoles of NaOH = C x 15

H3PO4 + NaOH ----------------------> NaH2PO4 + H2O

MV           C x 15

note : to solve this problem you have to give me molarity of H3PO4 & NaOH and H3PO4 volume also. otherwise we cannot solve

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