Question

How many grams of Na2CO3 (FM 105.99) should be mixed with 3.97 g of NaHCO3 (FM 84.01) to produce a 100 mL of buffer with pH of 10.03? (The Ka's of carbonic acid are Ka1 = 4.46 ✕ 10−7 and Ka2 = 4.69 ✕ 10−11.)

Answer 1

this is a buffer so

pH = pKa + log(CO3-2/HCO3-)

must be valid

we use 2nd ionization, since pH is nearest to KA2,

pKa2 = -log(Ka2) = -log(4.69*10^-11) = 10.328

10.03 = 10.328 + log ( CO3-2/HCO3-)

mol of HCO3- = mol of NaHCO3

mol of NaHCO3 = mass/MW = 3.97/84.01 = 0.047256 mol

10.03 = 10.328 + log ( CO3-2/0.047256 )

mol of Na2CO3 required

Na2CO3 = 0.047256 * 10^(10.03-10.328)

Na2CO3 = 0.023793 mol required

mass = mol*MW = 0.023793*105.99 = 2.5218 g of Na2CO3 required