3 Calculate the pH of 0375 L buffer solution made of a 018 M

3- Calculate the pH of 0.375 L buffer solution made of a 0.18 M Acetic acid HC2H3Oz(Ka = 1.8 x 10 s) and a 0.134 M Potassium acetate KC2H302 (do not use more than 3 digits beyond the decimal point) a) before adding anything b) after adding 0.010 mol Ba(OH)2 c) after adding 0.050 mol HCIO4

Solution

no of moles of CH3COOH   = molarity * volume in L

                                          = 0.18*0.375

                                           = 0.0675 moles

no of moles of CH3COOK   = molarity * volume in L

                                            = 0.134*0.375   = 0.05025 moles

PKa   = -logka

         = -log1.8*10^-5

         = 4.75

PH   = PKa + log[CH3COOK]/[CH3COOH]

        = 4.75 + log0.05025/0.0675

        = 4.75 -0.1281    = 4.62

b. after adding 0.01 mole Ba(OH)2

no of moles of CH3COOH after adding 0.01mole Ba(OH)2 = 0.0675-0.01   = 0.0575 moles

no of moles of CH3COOK after adding 0.01mole Ba(OH)2   = 0.05025+0.01 = 0.06025moles

PH     =   PKa + log[CH3COOK]/[CH3COOH]

         = 4.75 + log0.06025/0.0575

        = 4.75 +0.02029    = 4.77

c. after adding 0.05 mole HClO4

no of moles of CH3COOH after adding 0.05mole HClO4 = 0.0675+0.05 = 0.1175 moles

no of moles of CH3COOK after adding 0.05mole HClO4 = 0.05025-0.05 = 0.00025moles

PH     =   PKa + log[CH3COOK]/[CH3COOH]

            = 4.75 + log0.00025/0.1175

           = 4.75 -2.6720   = 2.08