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540 mL of .14M Formic Acid is combined with 330mL of .10M sodium hydroxide, calculate the...

540 mL of .14M Formic Acid is combined with 330mL of .10M sodium hydroxide, calculate the pH. Ka = 1.8e-4

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Expert Solution

no of moles of HCOOH   = molarity * volume in L

                                        = 0.14*0.54   = 0.0756moles

no of moles of NaOH    = molarity * volume in L

                                      = 0.1*0.33 = 0.033moles

                    HCOOH(aq) + NaOH(aq) -------------> HCOONa(aq) + H2O(l)

I                    0.0756               0.033                            0

C                   -0.033               -0.033                             0.033

E                  0.0426                 0                                   0.033

             Pka   = -logka

                       = -log1.8*10^-4

                       = 3.7447

            PH   = Pka + log[HCOONa]/[HCOOH]

                     = 3.7447 + log0.033/0.0426

                     = 3.7447 -0.11   = 3.6347 >>>>>answer

queen_honey_blossom answered 2 years ago
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