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What is the pH of an aqueous solution made by combining 44.32 mL of a 0.3616...

What is the pH of an aqueous solution made by combining 44.32 mL of a 0.3616 M ammonium chloride with 36.75 mL of a 0.3087 M solution of ammonia to which 4.002 mL of a 0.0516 M solution of NaOH was added?

Solutions

Expert Solution

no of moles of NH4Cl    = molarity * volume in L

                                      = 0.3616*0.04432   = 0.016 moles

no of moles of NH3    = molarity * volume in L

                                 = 0.3087*0.03675   = 0.0113 moles

PKb = 4.75

no of moles of NaOH = molarity * volume in L

                                    = 0.0516*0.004002   = 0.000207 moles

no of moles of NH4l after addition of NaOH = 0.016-0.000207   = 0.0158 moles

no of moles of NH3 after addition of NaOH = 0.0113+0.000207   = 0.011507 moles

POH   = PKb + log[NH4Cl]/[NH3]

          = 4.75 + log0.0158/0.011507

           = 4.75+0.137

             = 4.887

PH     = 14-POH

          = 14-4.887   = 9.113 >>>>answer


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