Question

A biochemist prepares two stocks of sodium phosphate buffer by titrating phosphoric acid with sodium hydroxide. She prepares a 0.10 M solution of sodium phosphate buffer at pH 2.15, and she prepares a 0.10 M solution of sodium phosphate buffer at pH 7.20. If she mixes 50. mL of the pH 2.15 buffer with 30. mL of the pH 7.20 buffer, what will be the pH of the resulting solution? Recall that the pKas of phosphoric acid are 2.15, 7.20, and 12.35.

Answer 1

buffer one:

H3PO4 ; NaH2PO4 are present

buffer two

NaH2PO4; Na2PO4 are present

in the final mix:

H3PO4, NaH2PO4, Na2HPO4 will be present

thre will be neutralizaiton of acids/bases

therefore

initially

V = 50 mL of buffer 1.

pH = pKa1 + log(NaH2PO4/H3PO4)

2.15 = 2.15 + log(NaH2PO4/H3PO4)

NaH2PO4/H3PO4 = 10^(2.15 -2.15 ) = 1

and we now

total concentraiton = 0.1 M of phosphates

V = 50 mL, mmol = M*V = 50*0.1 = 5 mmol

NaH2PO4/H3PO4 = 1

NaH2PO4 + H3PO4 = 5

solve

NaH2PO4 = H3PO4

NaH2PO4 + NaH2PO4 = 5

NaH2PO4 = 2.5 mmol

H3PO4 = 2.5 mmol

Now... second buffer:

pH = pKa2 + log(Na2HPO4/NaH2PO4)

7.20= 7.20+ log(NaH2PO4/NaH2PO4)

NaH2PO4/NaH2PO4= 10^(7.20-7.20)

NaH2PO4/NaH2PO4= 1

mmol of buffer = MV = 30*0.1 = 3

NaH2PO4/NaH2PO4= 1

NaH2PO4+ NaH2PO4= 3

Na2HPO4 = 1.5

NaH2PO4 = 1.5

Finally:

NaH2PO4 = 2.5 mmol + 1.5 mmol = 4 mmol

H3PO4 = 2.5 mmol

Na2HPO4 = 1.5 mmol

Note that

H3PO4; NaH2PO4; Na2HPO4

H3PO4-; H2PO4-; HPO4-2 are present

H3PO4- + HPO4-2 = H2PO4- + H2PO4-2

H3PO4- + HPO4-2 = 2H2PO4-

1.5 mmol of HP4-2 --> 1.5 mmol of H2PO4-

1.5 mmol of H3PO4 --> 1.5 mmol of H2PO4- formed, 1 mmol of H3PO4 left

so

H3PO4- = 1

H2PO4- = 4 + 1.5+1 = 6.5

HPO4-2 = 0

Now,

H3PO4- = 1

H2PO4- = 4 + 1.5+1 = 6.5

this is 1st ionizaiton

pH = pKa1 + log(H2PO4-/H3PO4)

pH = 2.15 + log(6.5/1)

pH = 2.96