[clevermartin]

In a experiment, I was asked to find the normality of Na²CO₃.
I was given 1.4g of Na₂CO₃ to dissolve in 250ml of H₂0 in a volumetric flask.
I wonder whether my calcultion is correct or not..

Equivalent weight (EW) = Molecular weight (MW) / Valencey. (Z)

Normality (N) = Equivalent weight (EW) / Volume of Solution (1000 ml.)

Meq = Actual weight (W) / Equivalent weight (EW)

Molecular weight of Na₂CO₃: 106
Valencey = 1 ? [(-2*3+4)/2]?
EW = 106/1 = 106
N = 106/1000 ? (Does it matter for dissolving the sodium carbonate in 250 ml volumetric flask?)

Thanks

[clevermartin]

And if we use H₂SO₄ to titrate against Na₂Co₃,
and find out that the molarity ratio between them is 1:1.
Is it their normality ration to be 1:1 also?

Thanks

[Borek]

http://www.chembuddy.com/?left=concentration&right=normality

[clevermartin]

Then I want to ask shall I count the number of sodium or carbonate for the valencey?

Thanks

[sjb]

You touch upon a good point, and part of the problem of simply asking for the normality of a solution without reference to a reaction that occurs.

What is the balanced reaction between Na₂CO₃ and H₂SO₄?

How would this differ if the reaction was between Na₂CO₃ and HCl?

S

[clevermartin]

Is it Na2CO3 will undergo two stages of reaction with H2SO4?
Having Carbonate turning into bicarbonate in first stage and all the bicarbonate turning into carbon dioxide in the second stage?
Overall reaction:
Na₂CO₃ + H₂SO₄ -> H₂O + CO₂ + Na₂SO₄

If we are using HCl, the reaction will stop after the first stage if the volume and molarity of acid and sodium carbonate added are the same?

But I still cannot get the point of how to choose between sodium or carbonate for valencey...

Please help.. Thanks