Question

In: Chemistry

you add 1 ml of 0.1M HCl to a solution containing 10ml of 0.2M acetic acid...

you add 1 ml of 0.1M HCl to a solution containing 10ml of 0.2M acetic acid (Ka = 1.8x10-5) and 10ml of 0.2 M sodium acetate. What is the resulting pH?

Solutions

Expert Solution

no of moles of CH3COOH = molarity * volume in L

                                           = 0.2*0.01 = 0.002moles

no of moles of CH3COONa = molarity * volume in L

                                           = 0.2*0.01 = 0.002 moles

no of moles of HCl    = molarity * volume in L

                                              = 0.1*0.001 = 0.0001moles

no of moles of CH3COOH after addition of 0.0001 moles of HCl   = 0.002+0.0001 = 0.0021 moles

no of moles of CH3COONa after addition of 0.0001 moles of HCl   = 0.002-0.0001   = 0.0019 moles

PKa   = -logKa

            = -log1.8*10^-5

             =4.74

PH   = Pka + log[CH3COONa]/[CH3COOH]

          = 4.74 + log0.0019/0.0021

          = 4.74 -0.04346

           = 4.6965 >>>>answer


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