Question

A solution of H2SO4 is standardized against a sample of

which has been previously found to contain 92.44? CaCO3 and

no other basic material. The sample weighing 0.7423 g is

treated with 42.42 mL of the acid and the excess acid then

requires 11.22 mL of NaOH solution. If 1.00 mL of the acid is

equivalent to 0.9976 mL of the NaOH, what is the N of each

solution?

Answer 1

to equivalent:

0.9976 NaOH - 1 ml H2SO4

11,22 NaOH - 11.247 ml H2SO4

0.7423 CaCO3 reacted with 42,42-11,247=31,173ml

nmoles CaCO3 is 0.007423 moles

Reactio:

CaCO3 +H2SO4=CaSO4+H2O+CO2↑

factor equivalemt is 1.

M(H2SO4)=0.2381/litr

N=M/Fequiv=0.2381 of acid

N of NaOH=0.23867