Design a buffer that has a pH of 8.52 using one of die weak base/conjugate acid

Question

Design a buffer that has a pH of 8.52 using one of die weak base/conjugate acid systems shown below. How many grams of the bromide salt of the conjugate acid must be combined with how many grams of the weak base, to produce 1.00 L of a buffer that is 1.00 M in the weak base? grams bromide salt of conjugate acid = grams weak base = Design a buffer that has a pH of 4.97 using one of the weak acid/conjugate base systems shown below. How many grains of the sodium salt of the weak acid must be combined with how many grams of tire sodium salt of its conjugate base, to produce 1.00 L of a buffer dial is 1.00 M in the weak base? grams sodium salt of weak acid = grams sodium salt of conjugate base =

Explanation / Answer

pH of buffer = pKa + log([base]/[acid])

a)

Molecular weight of weak base 1 (CH3NH2) = 31 g/mol
Molecular weight of conjugate acid bromide salt (CH3NH2HBr) = 31 + 1 + 79.9 = 111.9 g/mol
pKa = 10.62

pH = pKa + log([base]/[acid])

8.52 = 10.62 + log([base]/[acid])

log([base]/[acid]) = -2.1

([base]/[acid]) = 0.122

[base] = 1 M in 1 Litre

Moles of weak base = 1

Mass of weak base = 31 g

[acid] = 1 / 0.122 = 8.164 M

Moles of conjugate acid bromide salt = 8.164

Mass of conjugate acid bromide salt = 8.164 * 111.9 = 913.6 g

b)

Molecular weight of weak acid 1 (HC2O4-) = 89 g/mol
Molecular weight of conjugate base sodium salt (C2O4-Na) = 89 – 1 + 23 = 111 g/mol
pKa = 4.19

pH = pKa + log([base]/[acid])

4.97 = 4.19 + log([base]/[acid])

log([base]/[acid]) = 0.78

([base]/[acid]) = 2.17

[base] = 1 M in 1 Litre

Moles of weak base = 1

Mass of weak base = 111 g

[acid] = 1 / 2.17 = 0.46 M

Moles of conjugate acid = 0.46

Mass of conjugate acid = 0.46 * 89 = 40.95 g

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