Question

A `33.25*mL` volume of nitric acid was required for equivalence with a `0.425*g` mass of sodium carbonate. What is the concentration of the nitric acid?

Answer 1

We work out the number of moles of each reagent...............and calculate `[HNO_3]` to be a bit over `0.2*mol*L^-1`.

Explanation:

With all these problems, a stoichiometricall balanced equation that represents the reaction is a prerequisite.

`Na_2CO_3(aq) + 2HNO_3(aq) rarr 2NaNO_3(aq) + H_2O(l) + CO_2(g)uarr`

And then we work out the number of moles of the reagents.

`"Moles of sodium carbonate "=`

`(0.425*g)/(105.99*g*mol^-1)=4.01xx10^-3*mol`

Given that 2 equiv of nitric acid are required to neutralize each equiv sodium carbonate, we can work out the concentration of the nitric acid as:

`[HNO_3]=(2xx4.01xx10^-3*mol)/(33.25*mL)xx10^3*mL*L^-1~=`

`0.25*mol*L^-1`.