Answer to Question #112004 in Chemistry for Rena Perez

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