Question

38) A buffer is made that contains 0.50M acetic acid and 0.50M sodium acetate. To a 10o.0mL sample f the buffer 50.0 mL of 0.50M NaOH is added. What is the pH of the new solution? (K, 1.8x 10 for acetic acid) a) 5.11 b) 5.22 c) 5.33 d) 5.44 e) 5.55

Answer 1

no of moles of CH3COOH   = molarity * volume in L

                                             = 0.5*0.1   = 0.05 moles

no of moles of CH3COONa = molarity * volume in L

                                             = 0.5*0.1   = 0.05 moles

no of moles of NaOH            = molarity * volume in L

                                               = 0.5*0.05   = 0.025 moles

no of moles of CH3COOH after addition of 0.025moles of NaOH   = 0.05-0.025   = 0.025 moles

no of moles of CH3COONa after addition of 0.025 moles of NaOH = 0.05+0.025 = 0.075moles

Pka = -logKa

         = -log1.8*10^-5

          = 4.75

PH   = Pka + log[CH3COONa]/[CH3COOH]

         = 4.75 + log0.075/0.025

           = 4.74 + 0.477   = 5.22

b.5.22 >>>>answer