Answer to Question #147032 in General Chemistry for Tim

Question

When CaCO3 is used as a standard for a strong acid, it is necessary to dissolve it in an excess of
the acid and back titrate with NaOH solution. In such a standardization, a water suspension of
1.000g of CaCO3 was used. A volume of 49.85mL of HCl was added from the buret and after
warming the solution to remove any dissolved CO2, the solution required 6.32mL of NaOH to
reach an endpoint. If a separate 50.0mL-pipetful of the HCl required 48.95mL of the NaOH for
neutralization, what was the normality of the HCl and of the NaOH?

Answer 1

CaCO3 + 2HCl = CaCl2 + CO2 + H2O

1/100= 0.01 mol.

0.01 x 2 = 0.02 mol.

Cn= 0.02 x 1000/49.85 =0,4 N.

50 x 0.4/1000 = 0.02.

0.02 x 1000/ 48.95 = 0.4 N.