There is a typo error, we are provided with
2.00L of 1.00M KH2PO4 olution and 1.50L of 1.00M K2HPO4 solution.
Using H. H equation,
pH = pKa + log [anion] / [acid]
7.13 = 7.21 + log [anion] / [acid]
-0.08 =log [anion] / [acid]
or, 0.83176 = [anion] / [acid]
Ratio of concentrations in a solution, (450 ml), is the same as the ratio of moles in that volume.
arbitrarially choosing such that one to be added "X" ml of it , & the other becomes that its been added "250-X" ml
0.83176 = (450 ml - X ml)(1.00 M K2HPO4) / (X ml)(1.00 M KH2PO4)
0.83176 = (450 - X) / X
After solving X = 245.66 ml
Thus we mix 245.66 ml of 1.00 MKH2PO4 with 204 ml of 1.00 M K2HPO4
Thank you
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