Question

A saturated sodium carbonate solution at 100 degrees celsius contains 45.5 g of dissolved sodium carbonate per 100. mL of solution. The solubility product constant for sodium carbonate at this temp is

79.0

0.316

0.0790

36.8

316.

Answer 1

Molar mass of Na2CO3 = 2*MM(Na) + 1*MM(C) + 3*MM(O)

= 2*22.99 + 1*12.01 + 3*16.0

= 105.99 g/mol

Molar mass of Na2CO3= 105.99 g/mol

s = 45.5 g/100 mL = 455 g/L = 4.55*10^2 g/L

To covert it to mol/L, divide it by molar mass

s = 4.55*10^2 g/L / 105.99 g/mol

s = 4.293 g/mol

The salt dissolves as:

Na2CO3 <----> Na+ + 2 CO32-

   s 2s

Ksp = [Na+][CO32-]^2

Ksp = (s)*(2s)^2

Ksp = 4(s)^3

Ksp = 4(4.293)^3

Ksp = 316

Answer: 316