Answer to Question #319103 in General Chemistry for Green

Question

A 0.205 g sample of CaCO3 (Mr = 100.1 g/mol) is added to a flask along with 7.50 mL of 2.00 M HCl. CaCO3(aq) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)Enough water is then added to make a 125.0 mL solution. A 10.00 mL aliquot of this solution is taken and titrated with 0.058 M NaOH. NaOH(aq) + HCl(aq) → H2O(l) + NaCl(aq)How many mL of NaOH are used?

Answer 1

Mole ratio of CaCO₃ : HCl = 1:2

Moles of CaCO₃ = 0.205÷100.1 = 0.002048

Moles of HCl = 2×7.5÷1000 = 0.015

Moles of excess HCl= 0.015-0.002048×2

=0.010904

0.010904 moles of HCl will be in 125ml

Moles of HCl in 10.00ml= 0.010904×10÷125 = 0.00087232

Mole ratio NaOH:HCl = 1:1

Moles of NaOH required = 0.00087232

Volume of NaOH= 0.00087232×1000÷0.058

=15.04 ml