Question

What would be the pH of a solution comprised of 50 mL of 0.1 M H3PO4 and 25 mL of 0.1 M KH2PO4?

H3PO4 H2PO4− + H+ pKa1 = 2.15

H2PO4− HPO42− + H+ pKa2 = 7.20

HPO42− PO43− + H+ pKa3 = 12.37

Please show answer in detail

Answer 1

no of moles of H3PO4 = molarity * volume in L

                                     = 0.1*0.05   = 0.005 moles

no of moles of KH2PO4 = molarity * volume in L

                                        = 0.1*0.025   = 0.0025 moles

PH = Pka + log[KH2PO4/[H3PO4]

       = 2.15 + log0.0025/0.005

    = 2.15-0.3010

    = 1.849 >>>>>answer