Question

How many grams of Na₂CO₃ (FM 105.99) should be mixed with 6.41 g of NaHCO₃ (FM 84.01) to produce a 100 mL of buffer with pH of 10.14? (The K_a's of carbonic acid are K_a1 = 4.46 ✕ 10⁻⁷ and K_a2 = 4.69 ✕ 10⁻¹¹.)
NOTE: Use three significant figures in answer.

Answer 1

we know that

moles = mass / molar mass

so

moles of NaHC03 = 6.41 / 84.01

moles of NaHC03 = 0.07630044

we know that

for a buffer solution

pH = pKa + log [salt / acid ]

in this case

pH = pKa2 + log [ Na2C03 / NaHC03]

now

we know that

pKa2 = -log Ka2

so

pH = -log Ka2 + log [ Na2C03 / NaHC03]

10.14 = -log 4.69 x 10-11 + log [ Na2C03 / NaHC03]

[ Na2C03 / NaHC03] = 0.6474

now

we know that

concentration = moles / volume

as the final volume is same

ratio of concentrations = ratio of moles

so

moles of Na2C03 / moles of NaHC03 = 0.6474

so

moles of Na2C03 / 0.07630044 = 0.6474

moles of Na2C03 = 0.0494

now

mass = moles x molar mass

so

mass of Na2C03 = 0.0494 x 105.99

mass of Na2C03 = 5.236

so

5.24 grams of Na2C03 is required