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Determine the volume in mL of 0.15 M NaOH(aq) needed to reach the equivalence (stoichiometric) point...

Determine the volume in mL of 0.15 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 48 mL of 0.12 M HF(aq). The Ka of HF is 7.4 x 10-4.

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Answer #1

      no of moles of HF = molarity * volume in L

                                    = 0.12*0.048   = 0.00576moles

         M1V1/n1    =   M2V2/n2

                V2 = M1V1n2/n1M2

                        =0.12*48*1/1*0.15   = 38.4ml

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