Question

Exercise 16.50

For each of the following solutions, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH.

Part A

For 300.0 mL of pure water, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH.

Express your answers using two decimal places separated by a comma.

pHinitial, pHfinal =

7.00,12.82

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Correct

Part B

For 300.0 mL of a buffer solution that is 0.210 M in HCHO2 and 0.310 M in KCHO2, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Ka=1.8⋅10−4).

Express your answers using two decimal places separated by a comma.

pHinitial, pHfinal =

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Part C

For 300.0 mL of a buffer solution that is 0.2551 M in CH3CH2NH2 and 0.2306 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.0200 mol of NaOH(Kb=5.6⋅10−4).

Express your answers using two decimal places separated by a comma.

How to do B and C

Answer 1

part B

pH of buffer = pka + log(salt/acid)

pka of HCHO2 = -log(1.8*10^-4) = 3.745

no of mol of salt (KCHO2) = 0.31*0.3 = 0.093 mole

no of mol of acid(HCHO2) = 0.21*0.3 = 0.063 mole

pH = 3.745 + log(0.093/0.063)

initial pH   = 3.91

after NaOH added

pH of buffer = pka + log(salt+NaOH/acid-NaOH)

no of mole of NaOH added = 0.02 mol

    = 3.745 + log((0.093+0.02)/(0.063-0.02))

final pH = 4.165

part C

pH of basic buffer = 14-(pkb + log(salt/base))

pkb of CH3CH2NH2 = -log(5.6*10^-4) = 3.25

no of mol of salt (CH3CH2NH3+) = 0.2306*0.3 = 0.0692 mole

no of mol of base(CH3CH2NH2) = 0.2551*0.3 = 0.0765 mole

pH = 14 - (3.25 + log(0.0692/0.0765)

initial pH   = 10.8

after NaOH added

pH of buffer = pka + log(salt-NaOH/base+NaOH)

no of mole of NaOH added = 0.02 mol

    = 14 - (3.25 + log((0.0692-0.02)/(0.0765+0.02))

     = 11.04