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

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Answer 1 · 2016-11-17T18:32:11

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

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#.

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?