Question

A 2.500 grams sample of a mixture of sodium carbonate (Na2CO3) and sodium chloride (NaCl) is dissolved in 25.00 mL of 0.798 M HCl (hydrochloric acid). Some acid will remain after the treatment of the sample. If 28.7 mL of 0.108 M NaOH is required to titrate the excess HCI then how many moles of sodium carbonate were present in the original sample? Hint: The moles of Na2CO3 present is equal to one- half the number of H30* moles and the reaction between Na2CO3 and H30+ generates H2CO3 which immediately decomposes to H20 and CO2 Na2CO3 + 2H30+ → H2CO3 + H2O + 3H20 + CO2

Answer 1

The balanced equations are Na2CO3 + 2 H2O+ --------> 3 H2O + CO, + 2 Na+ NaOH + H2O+ ----------> 2 H2O + Nat Number of moles of HCl added initially=M*V = 0.798 M * 25.00 mL = 19.95 mmol Number of moles of NaOH used to titrate excess HCI = M*V = 0.108 M * 28.7 mL = 3.0996 mmol Since number of moles of NaOH = number of moles of HCI Number of moles of HCl present excess (remain unreacted after reaction with Na2CO3 ) = 3.0996 mmol Number of moles of HCl that react with Na2CO3 = 19.95 – 3.0996 = 16.8504 mmol According to balanced equation, 2 moles of HCl reacts with 1 mole of Na Coz So, 16.8504 mmol of HCl would react with 16.8504 mmol * 12 = 8.4252 mmol of Na, Coz Number of moles of Na2CO3 = 8.4252 mmol = 0.008425 mol