Answer to Question #174791 in Chemistry for Dan

Question

A 0.3749 g soda ash sample is analyzed by titrating sodium carbonate with the standard 0.2388M HCl solution, requiring 49.38ml. The reaction is:

          CO₃^2-  + 2H⁺     →   H₂O +CO₂

Answer 1

soda ash = sodium carbonate = Na₂CO₃

Solution:

The reaction is:

CO₃^2- + 2H⁺ → H₂O + CO₂

According to the reaction above: n(CO₃^2-) = n(H⁺)/2

HCl → H⁺ + Cl^-

n(H⁺) = n(HCl) = C_M(HCl) × V(HCl) = (0.2388 M) × (0.04938 L) = 0.011792 mol

n(CO₃^2-) = n(H⁺) / 2 = 0.011792 mol / 2 = 0.005896 mol

Na₂CO₃ → 2Na⁺ + CO₃^2-

n(Na₂CO₃) = n(CO₃^2-) = 0.005896 mol

Mass of Na₂CO₃ = Moles of Na₂CO₃ × Molar mass of Na₂CO₃

The molar mass of Na₂CO₃ is 105.9888 g mol^-1.

Hence,

Mass of Na₂CO₃ = (0.005896 mol) × (105.9888 g mol^-1) = 0.6249 g

Mass of Na₂CO₃ = 0.6249 g

%Na₂CO₃ = (Mass of Na₂CO₃ / Mass of soda ash sample) × 100%

However,

Mass of Na₂CO₃ > Mass of soda ash sample

0.6249 g > 0.3749 g

Perhaps there was an error in the problem statement!

Since the percent sodium carbonate in the sample cannot be more than 100%.