Question

How many grams of Na2CO3 (MW = 105.9885 g/mole) need to be dissolved in a 0.75 M NaHCO3 solution in order to create a buffer at a pH 9.75 that has a total volume of 500.0 mL? (The answer is 13g Na2CO3 but show step by step on how to get to the answer)

Answer 1

Na₂CO₃ -----> Na²⁺ + CO₃^2-

[Na₂CO₃] dissolved = [CO₃^2-] in the solution

Let concentration of CO₃^2- obtained by dissolving Na₂CO₃ be x M.

HCO₃^- ------> H⁺ + CO₃^2-

From th balanced equation, it is clear that, at equilibrium

[H⁺] = [CO₃^2-]

At equilibrium pH = 9.75 = -log[H⁺]

hence, [H⁺] = antilog(-9.75) = 1.78 x 10^-10

so, [CO₃^2-] from acid = 1.78 x 10^-10

Total concentration of carbonate in the solution is

[CO₃^2-] = 1.78 x 10^-10 + x

at equilibrium, undissociated bicarbonate ion concentration is (0.75 - 1.78 x 10^-10) 0.75

For bicarbonate, Ka = 4.8 x 10^-11

Therefore,

solving the above equation, we get:

x = 0.202

Hence, concentration of Na₂CO₃should be 0.202 M

Then, no. of moles of Na₂CO₃ in 500 mL (=0.500 L) of the solution = (0.202 mol L^-1)(0.500 L) = 0.101 mol

mass of 0.101 mol Na₂CO₃ = (0.101 mol)(105.9885 g/mol) = 10.7 g

(this value may be les than 13 due to some approximations done)

amount of sodium carbonate needed is 10.7 g.