Question

what is the ph when 25.00 no of .20 M CH3COOH has been treated with 40.0 no of .10 M NaOH?

Answer 1

no of moles of CH3COOH = molarity * volume in L

                                             = 0.2*0.025 = 0.005 moles

no of moles of NaOH          = molarity * volume in L

                                            = 0.1*0.04 = 0.004 moles

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

I               0.005             0.004                                  0           

C            -0.004            -0.004                                  0.004

E             0.001             0                                         0.004

            [CH3COONa]   = 0.004 moles

            [CH3COOH]    = 0.001 moles

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

                 = 4.75 + log0.004/0.001

                = 4.75 + 0.6020   = 5.352 >>>>>>answer