Question

For each of the following solutions, calculate the initial pHand the final pH after adding 0.005 mol of NaOH.

260.0 mL of a buffer solution that is 0.225 M in HCHO2 and 0.315 M in KCHO 260.0 mL of pure water 260.0 mL of a buffer solution that is 0.275 M in CH3CH2NH2 and 0.245 M in CH3CH2NH3Cl

Answer 1

1) initially

pH = pKa + log [HCHO2] / [CHO2^-]

pKa = HCHO2 = 3.80

pH = 3.80 + log [0.315] / [0.225]

pH = 3.95

after 0.005 / 0.26 = 0.0192 M NaOH added

pH = 3.80 + log [0.315 + 0.0192] / [0.225 - 0.0192]

pH = 3.80 + log [0.3342] / [0.2058]

pH = 4.01

2) for pure water

pH = 7.0

after 0.005 / 0.26 = 0.0192 M NaOH added

pOH = - log [OH^-]

as NaOH is strong base

pOH = - lg [0.0192]

pOH = 1.72

pH = 14 - 1.72

pH = 12.28

3) initially

pOH = pKb + log [CH3CH2NH3Cl] / [CH3CH2NH2]

pKb of CH3CH2NH2 = 3.20

pOH = 3.20 + log [0.245] / [0.275]

pOH = 3.15

pH = 14 - 3.15

pH = 10.85

after0.005 / 0.26 = 0.0192 MNaOH added

pOH = 3.20 + log [0.245 - 0.0192] / [0.275 + 0.0192]

pOH = 3.20 + log [0.2258] / [0.2942]

pOH = 3.08

pH = 14 - 3.08

pH = 10.92