Question

Calculate the pH of a buffer that is 0.10 M in KH2PO4 and 0.050 M in Na2HPO4. What is the pH after adding 5.0 mL of 0.20 M HCl to 0.10 L of this buffer. The Ka value for H2PO4– is 6.32 × 10–8 or a pKa of 7.199

Answer 1

pH = pKa + log (base)/(acid)
base is 0.050M Na₂HPO₄
acid is 0.10 M KH₂PO₄

pKa = 7.199

pH = 7.199 + log (0.05)/(0.1)

pH = 6.89

100.0 mL x 0.050 = 5 mmols base Na₂HPO₄
100.0 mL x 0.10 = 10 mmols acid KH₂PO₄
added 5 mL x 0.20 M HCl = 1.0 mmols

. HPO₄^-   + H⁺ ==> H₂PO₄^-

                              Initial                                   5             0             10

                              Add                                                     1.0

                              Change -1            -1            +1

                              Equilibrium                        4             0             11

pH = pKa2 + log (base)/(acid)

base is 4 mmols Na₂HPO₄

acid is 11 mmol KH₂PO₄

pH = 7.199 + log 4/11

pH = 6.76