Question

Item 1 You may want to reference (Pages 781 - 791) section 17.2 while completing this problem. Calculate the pH of the solution that results from each of the following mixtures. Part A 150.0 mL of 0.24 M HF with 230,0 mL of 0.32 M NaF Express your answer using two decimal places. 290 ADO ? pH = Submit Request Answer Part B 165.0 mL of 0.12 M C,H, NH, with 265,0 mL of 0.20 M C H NH,CI Express your answer using two decimal places. EVO AXO + O ? pH =

Answer 1

no of moles of HF = molarity *volume in L

                           = 0.24*0.15   = 0.036moles

no of moles of NaF = molarity *volume in L

                             = 0.32*0.23   = 0.0736moles

Pka of HF   = 3.14

PH   = PKa + log[NaF]/[HF]

       = 3.14 + log0.0736/0.036

       = 3.14 + 0.3105

      = 3.45 >>>>answer

part-B

no of moles of C2H5NH2   = molarity * volume in L

                                      = 0.12*0.165   = 0.0198moles

no of moles of C2H5NH3Cl = molarity * volume in L

                                        = 0.20*0.265   = 0.053moles

Pkb of C2H5NH2    = 3.2

POH    = Pkb + log[C2H5NH3Cl]/[C2H5NH2]

         = 3.2 + log0.053/0.0198

        = 3.2 + 0.4276

        = 3.6276

PH   = 14-POH

       = 14-3.6276

       = 10.37