Question

How many grams of sodium formate (68.01 g/mol) must be added to 2.0L of 0.10 M formic acid (Ka=1.77 x 10^-4) to form a buffer whose pH is 3.50? (Assume that the addition of the salt does not change the volume of the solution).

Answer 1

no of moles of HCOOH     = molarity*volume in L

                                            = 0.1*2   = 0.2 moles

PH     =   3.5

PKa   = -logKa

           = -log1.77*10^-4

           = 3.752

PH    = Pka + log[HCOONa]/[HCOOH]

3.5   = 3.752 + log[HCOONa]/0.2

log[HCOONa]/0.2   = 3.5-3.752

log[HCOONa]/0.2 = -0.252

[HCOONa]/0.2        = 10^-0.252

[HCOONa]/0.2        = 0.5597

[HCOONa]               = 0.5597*0.2

[HCOONa]                = 0.11194 moles

mass of HCOONa   = no of moles * gram molar mass                      

                                 = 0.11194*68.01

                                  = 7.613 g >>>>>answer