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Step1 Concentration of formic acid = .200x20 = 4 milimoles Step2 Concentration
of NaOH= 25 x.200 = 5 milimoles Step3 Net concentration of NaOH = 1 milimole = 1 x10^-4
Mole Step4 [H+][OH-] = Kw= 10^-14 ; [H+][1x10^-4]=10^-14 Step5 [H+] = 1x10^-10 Step6
pH=-log[H+]=-log[1x10^-10] =10
Solution
Step1 Concentration of formic acid = .200x20 = 4 milimoles Step2 Concentration
of NaOH= 25 x.200 = 5 milimoles Step3 Net concentration of NaOH = 1 milimole = 1 x10^-4
Mole Step4 [H+][OH-] = Kw= 10^-14 ; [H+][1x10^-4]=10^-14 Step5 [H+] = 1x10^-10 Step6
pH=-log[H+]=-log[1x10^-10] =10

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Step1 Concentration of formic acid = .200x20 = 4 .pdf

  • 1. Step1 Concentration of formic acid = .200x20 = 4 milimoles Step2 Concentration of NaOH= 25 x.200 = 5 milimoles Step3 Net concentration of NaOH = 1 milimole = 1 x10^-4 Mole Step4 [H+][OH-] = Kw= 10^-14 ; [H+][1x10^-4]=10^-14 Step5 [H+] = 1x10^-10 Step6 pH=-log[H+]=-log[1x10^-10] =10 Solution Step1 Concentration of formic acid = .200x20 = 4 milimoles Step2 Concentration of NaOH= 25 x.200 = 5 milimoles Step3 Net concentration of NaOH = 1 milimole = 1 x10^-4 Mole Step4 [H+][OH-] = Kw= 10^-14 ; [H+][1x10^-4]=10^-14 Step5 [H+] = 1x10^-10 Step6 pH=-log[H+]=-log[1x10^-10] =10