Question

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH. Part A For 220.0 mL of a buffer solution that is 0.235 M in HCHO 2 and 0.290 M in KCHO 2 , calculate the initial pH and the final pH after adding 0.0200 mol of NaOH( K a =1.8⋅ 10 −4 ) . Part B For 220.0 mL of a buffer solution that is 0.2710 M in CH 3 CH 2 NH 2 and 0.2510 M in CH 3 CH 2 NH 3 Cl , calculate the initial pH and the final pH after adding 0.0200 mol of NaOH( K b =5.6⋅ 10 −4 ) .

Answer 1

Part A .

pKa = -log Ka = -log (1.8 x 10^-4)

       = 3.74

pH = pKa + log [KCHO2 / HCHO2]

      = 3.74 + log (0.290 / 0.235)

     = 3.83

initial pH = 3.83

moles of KCHO2 = 0.290 x 220 / 1000 = 0.0638

moles of HCOOH = 0.0517

pH = pKa + log [salt + C / acid - C]

     = 3.74 + log [0.0638 + 0.02 / 0.0517 - 0.02]

    = 4.16

final pH = 4.16

Part B .

pKb = -log Kb = -log (=5.6 x 10^ −4)

       = 3.25

pOH = pKb + log [salt / base]

        = 3.25 + log [0.2510 / 0.2710]

        = 3.22

initial pH = 10.78

moles of base = 0.05962

moles of salt = 0.05522

pOH = pKb + log [salt - C / base + C]

       = 3.25 + log [0.05522 - 0.02 / 0.05962 + 0.02]

      = 2.90

pH = 11.10

final pH = 11.10