Question

Gr. 12 Chemistry (Ontario Curriculum)
Q. How many grams of Na₂CO₃ ⋅ 10H₂O have to be dissolved in 1.00L of water at 25°C to have a
solution with a pH of 11.70? [35g] *The final answer is in Bold, but I need solution steps*

Answer 1

pH = 11.7 ,

pOH = 14 -pH = 14 - 11.7 = 2.3

[OH-] = 10^- pOH = 10^- 2.3 = 0.005 M

Now this OH- is produced by Na2CO3 i.e CO3^2- ions

The equation is given by   CO3^2- (aq) + H2O (l) <----> HCO3- (aq) + OH- (aq)

at equilibrium we have [CO32-] = A-X   where A is initial Molarity of CO3^2- ,

[HCO3-] =[OH-] = X = 0.005, hence [CO32-] = A-0.005

Kb of CO3^2- = Kw / ( Ka of HCO3-) = 10^-14 / ( 4.8 x 10^-11) = 0.0002

Kb = [HCO3-] [OH-] / [CO3^2-]

0.0002 = (0.005) ( 0.005) / ( A-0.005)    ( we approximate A-0.005 as nearly A , to simplify)

A = 0.125 mol/L = initail Molarity of Na2CO3

volume = 1L , hence moles = M x V = 0.125 mol/L x 1 L = 0.125 mol

Mass of Na2CO3 = moles x molar mass of Na2CO3.10H2O

         = 0.125 mol x 286 g/mol = 35.75 g

( Note I used Ka , Kb values from web, if data table given using appropriate Kb for Co3^2- we can get 35 g)