Question #33729

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

Expert's answer

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 Na_{2}CO_{3} having 106 g / mole & withthe conj acid NaHCO_{3} 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 [Na_{2}CO_{3}]/[NaHCO_{3}]

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 NaHCO_{3}

SECOND ANSWER: 6.0 grams of Na_{2}CO_{3}

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

-------------------------

since it is in one litre, Molarity = grams / MolarMass & with the conj base Na

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

which became:

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

becomes:

9.65 = 10.25 + log [Na

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 NaHCO

SECOND ANSWER: 6.0 grams of Na

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