Question

calculate the ph of a buffer that is 0.120M in NaHCO3 and 0.190M in Na2CO3

Answer 1

pH = pK_a + log[Base]/[Acid]

9.65 = 10.25 + log [Base]/[Acid]
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since it is in one litre, Molarity = grams / MolarMass & with the conj base Na₂CO₃ having 106 g / mole & withthe conj acid NaHCO₃ having 84 g / mole


9.65 = 10.25 + log [Base]/[Acid]

which became:
9.65 = 10.25 + log [Base]/[Acid]

becomes:
9.65 = 10.25 + log [Na₂CO₃]/[NaHCO₃]

becomes:
9.65 = 10.25 + log [25g -X / 106] / [X / 84]

rearranges into :
9.65 - 10.25 = log [25g -X / 106] / [X / 84]

- 0.6 = log [25g -X / 106] / [X / 84]

out of logs:
0.251 = [25g -X / 106] / [X / 84]

rearranges into:
(0.251) [X / 84] = [25g -X / 106]

or:
(0.251) (X) / (84) = (25g -X) / (106)

cross multiply:
(0.251) (X) (106) = (25g -X) (84)

expands into:
26.62 X = 2100 - 84 X

rearranges into
110.62 X = 2100

FIRST ANSWER: 19 grams of NaHCO₃
SECOND ANSWER: 6.0 grams of Na₂CO₃