Question

Sodium carbonate is a reagent that may be used to standardize acids in the same way that KHP was used in this experiment. In such a standardization it was found that a 0.512 g sample of sodium carbonate required 25.3 mL of a sulfuric acid solution to reach the end point for the reaction. Na2CO3(aq) + H2SO4(aq)  H2O(l) + CO2(g) + Na2SO4(aq) What is the molarity of the H2SO4?

Answer 1

Solution: The reaction is given as:

Na2CO3 + H2SO4 = H2O + CO2 + Na2SO4

From the reaction between Na2CO3 and H2SO4, it can be seen that 1 mol of Na2CO3 required 1 mol of H2SO4.

Since. Number of moles = mass / molar mass

Therefore, number of moles of Na2CO3 in 0.512 g of sample

= 0.512 g/ 105.9884g/mol = 0.00483 mol

Since,

Number of moles of Na2CO3 = Number of moles of H2SO4

Therefore, number of moles of H2SO4 will be =0.00483 mol

Volume of H2SO4 = 25.3 ml = 0.0253 L

Molarity (M) is calculated as,

M = Number of moles / Volume in L

M= 0.00483 mol / 0.0253 L = 0.19 mol/L = 0.19 M