Question

1.725 g of a metal carbonate is mixed with 300 mL of N/10 HCl.
10 mL of sodium hydroxide were required to neutralise
N/2 excess of the acid. Calculate the equivalent mass of the metal carbonate.
[Ans. 69]​

Answer 1

Nv = N'v'

let excess of the acid be v'. hnce , 10 x 1/2 = 1/10 xv'

hence v' = 50ml. hence this means that remaining volume of hcl reacts with the metal carbonate

which is 300-50 ml of the acid. = 250 ml.

again, NxV=N'xV'

(wx1000/Ewt) x vol x vol = N' x V'

[volume cancels out] hence we get, after substituting values ----------- 1.725x1000/Ewt = 250 x 1/10

hence Ewt = 1.725x1000/25 = 69 amu.

hence the equivalent wt of the metal carbonate is 69 amu