Question

How many grams of Na2CO3 (FM 105.99) should be mixed with 4.61 g of NaHCO3 (FM 84.01) to produce a 100 mL of buffer with pH of 10.44? (The Ka's of carbonic acid are Ka1 = 4.46 ✕ 10−7 and Ka2 = 4.69 ✕ 10−11.)

Answer 1

we know that

moles = mass / molar mass

so

moles of NaHC03 = 4.61 / 84.01

moles of NaHC03 = 0.0548744

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]

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

[ Na2C03 / NaHC03] = 1.2917

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 = 1.2917

so

moles of Na2C03 / 0.0548744 = 1.2917

moles of Na2C03 = 0.070883

now

mass = moles x molar mass

so

mass of Na2C03 = 0.070883 x 105.99

mass of Na2C03 = 7.513

so

7.513 grams of Na2C03 is required