Question

To make 1.0 L of buffered solution of a 10.0 pH and a total [ ] of carbonate and hydrocarbonate is 0.10 M, how many grams of Na₂CO3 and NaHCO₃ are needed?

Carbonic acid K_a1 = 4.3 * 10^-7 and K_a2 = 5.6 * 10^-11

Answer 1

We have, pKa = - log Ka

$\therefore$ pKa = - log ( 5.6 $\times$ 10 ^-11) = 10.25

pH of buffer solution is calculated by using Henderson's equation.

pH = pKa + log [ Na₂CO₃ ] / [ NaHCO₃ ]

10.0 = 10.25 + log [ Na₂CO₃ ] / [ NaHCO₃ ]

log [ Na₂CO₃ ] / [ NaHCO₃ ] = 10.0 - 10.25 = - 0.25

[ Na₂CO₃ ] / [ NaHCO₃ ] = 10 ^-0.25 = 0.562

[ Na₂CO₃ ] / [ NaHCO₃ ] = 0.562

i e [ Na₂CO₃ ] = 0.562 [ NaHCO₃ ] ---------------------->(1)

We have, [ Na₂CO₃ ] + [ NaHCO₃ ] = 0.10 M

From equation 1 , we can write

0.562 [ NaHCO₃ ] + [ NaHCO₃ ] = 0.10

1.562 [ NaHCO₃ ] = 0.10

[ NaHCO₃ ] = 0.10 / 1.562

[ NaHCO₃ ] =0.0640 M

$\therefore$ [ Na₂CO₃ ] = 0.562 [ NaHCO₃ ] becomes [ Na₂CO₃ ] = 0.562 ( 0.0640 ) = 0.0360 M

Volume of buffer solution = 1.0 L

We know that, Molarity = No. of moles of Solute / volume of solution in L

$\therefore$ [ NaHCO₃ ] = No. of moles of  NaHCO₃ / volume of solution in L

$\therefore$ 0.0640 mol / L = No. of moles of  NaHCO₃ / 1.0 L

No. of moles of  NaHCO₃ = 0.0640 mol / L $\times$ 1.0 L = 0.0640 mol

No. of moles = Mass / Molar mass

$\therefore$ Mass of NaHCO₃ = 0.0640 mol ( 84.00 g/ mol ) = 5.376 g

Calculation for Na₂CO₃

[ Na₂CO₃ ] =No. of moles of Na₂CO₃ / volume of solution in L

$\therefore$ 0.0360 mol / L = No. of moles of Na₂CO₃ / 1.0 L

No. of moles of Na₂CO₃ = 0.0360 mol / L $\times$ 1.0 L = 0.0360 mol

We know that, No. of moles = Mass / Molar mass

$\therefore$ Mass of Na₂CO₃= 0.0360 mol ( 105.99 g/ mol ) = 3.816 g

ANSWER : Mass of NaHCO₃ = 5.376 g

Mass of Na₂CO₃= 3.816 g