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Calculate the pH of a 1.0 L solution that contains 0.50 M NH_3 and 0.40 M NH_4Cl

ID: 1021981 • Letter: C

Question

Calculate the pH of a 1.0 L solution that contains 0.50 M NH_3 and 0.40 M NH_4Cl. The K_b of NH_3 is 1.8 times 10^-5. A 1.0-L solution contains 0.50 M NH_3 and 0.40 M NH_4Cl. Calculate the pH of the solution after the addition of 100.0 mL of 1.0 M HCl. The K_b of NH_3 is 1.8 times 10^-5. 25.0 mL of 0.10 M formic acid (HCHO_2) is titrated with 0.10 M NaOH solution. Calculate the pH of the acid solution before any base has been added. The K_a of formic acid is 1.7 times 10^-4. 25.0 mL of 0.10 M formic acid (HCHO_2) is titrated with 0.10 M NaOH solution. Calculate the pH of the acid solution at the equivalence point of the titration. The k_a of formic acid is 1.7 times 10^-4.

Explanation / Answer

32. POH = PKb + log[NH4Cl]/[NH3]

no of moles of NH4Cl   = molarity * volume in L

                                      = 0.4*1 = 0.4 moles

no of moles of NH3     = molarity * volume in L

                                   = 0.5*1   = 0.5moles

                        Pkb = -logKb

                                = -log1.8*10-5

                                       = 4.75

                POH    = Pkb + log[NH4Cl]/[NH3]

                           = 4.75 + log0.4/0.5

                          = 4.75-0.0969

                          = 4.6531

                  PH   = 14-POH

                         = 14-4.6531

                         = 9.3469 >>>>> answer

33. By the addition of 100ml of 1 M HCl

       no of moles of HCl   = 1*0.1 = 0.1 moles

   no of moles of NH4Cl   = 0.4+0.1 = 0.5 moles

   no of moles of NH3      = 0.5-0.1 = 0.4 moles

                           POH   = Pkb + log[NH4Cl]/[NH3]

                                    = 4.75 + log0.5/0.4

                                    = 4.75 + 0.09691

                                     = 4.8469

                            PH   = 14-POH

                                    = 14-4.8469 = 9.1531 >>>> answer

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