Question

Use the Henderson–Hasselbalch equation to calculate the pH of each solution:

Part A

a solution that is 16.5 g of HF and 22.0 g of NaF in 125 mL of solution

Part B

a solution that contains 1.23% C2H5NH2 by mass and 1.30% C2H5NH3Br by mass

Part C

a solution that is 15.0 g of HC2H3O2 and 15.0 g of NaC2H3O2 in 150.0 mL of solution

Answer 1

A. no of moles of HF = W/G.M.Wt

                                   = 16.5/20   = 0.825moles

molarity of HF            = no of moles /volume in L

                                  = 0.825/0.125   = 6.6 M

no of moles of NaF    = 22/42 = 0.524moles

molarity of NaF          = 0.524/0.125 = 4.192

          PH               = Pka + log[NaF]/[HF]

                               = 3.17 + log4.192/6.6

                               = 3.17-0.1971 = 2.9729

part-B

     no of moles of CH3NH2 = 1.23/31 = 0.03967 moles

     no of moles of CH3NH3Br   = 1.3/112 = 0.0116moles

       POH = Pkb + log[CH3NH3Br]/[CH3NH2]

                 = 3.3665 + log0.0116/0.03967

                 = 3.3665-0.534 = 2.8325

     PH   = 14-POH

           = 14-2.8325   = 11.1675