A buffer solution contains 0.233 M ammonium chloride and 0.321 M ammonia. If 0.0

Question

A buffer solution contains 0.233 M ammonium chloride and 0.321 M ammonia. If 0.0188 moles of perchloric acid are added to 150 mL of this buffer, what is the pH of the resulting solution ? (Assume that the volume does not change upon adding perchloric acid)

pH =

A buffer solution contains 0.338 M NaH2PO4 and 0.394 M Na2HPO4.

If 0.0497 moles of perchloric acid are added to 225 mL of this buffer, what is the pH of the resulting solution ?
(Assume that the volume does not change upon adding perchloric acid)

pH =

Explanation / Answer

Q1.

The buffer equation

pOH = pKb + log(NH4+(NH3)

so

pKb = 4.75 for NH3

mmol of NH3 = MV = 0.321*150 = 48.15 mmol

mmol of NH4+ = MV = 0.233*150 = 34.95 mmol

after adding:

0.0188 mol of H+ = 0.0188*10^3 = 18.8 mmol

then

mmol of NH3 = 48.15- 18.8 = 29.35

mmol of NH4+ = 34.95+18.8 = 53.75

so

from

pOH = pKb + log(NH4+(NH3)

pOH = 4.75 + log(53.75/29.35)

pOH = 5.0127

pH = 14-5.0127 = 8.9873

Q2.

For

H2PO4- and HPO4-2

the ionization is pKa2, for pshophoric acid

so

pH = pKa2 + log(HPO4-2/H2PO4-)

pKA2 = 7.21 for phosphoric acid

initially

mmol of H2PO4- = 0.338*225 =76.05

mmol of HPO4-2 = 0.394*225 = 88.65

after adding

0.0497 mol = 49.7 mmol of H+

mmol of H2PO4- = 76.05 +49.7 = 125.75

mmol of HPO4-2 = 88.65-49.7 = 38.95

substitue in pH

pH = 7.21 + log(38.95/125.75) = 6.7009

pH = 6.7009

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