Question

Q7 A 40.00ml sample of .10M weak acid with ka of 1.8x10^-5 is titrated with a 0.10 M strong base. What is the pH after 20.00mL of base has been added?

9.26

7.00

1.8x10^-5

^4.74

Answer 1

let week acid is CH3COOH

let strong base is NaOH

no of moles of acid CH3COOH = molarity * volume in L

                                        = 0.1*0.04   = 0.004 moles

no of moles of strong base NaOH = molarity * volume in L

                                                        = 0.1*0.02 = 0.002moles

                   CH3COOH + NaOH --------------> CH3COONa + H2O

I                  0.004              0.002                           0               

C              -0.002             -0.002                         0.002

E              0.002                0                                0.002

      Pka = -logKa

                = -log1.8^10^-5

              = 4.74

   PH   = Pka + log[CH3COONa]/[CH3COOH]

        = 4.754 + log0.002/0.002

     = 4.74 + log1

      = 4.74 + 0

    = 4.74 >>>>answer