# 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?

Nov 17, 2016

We work out the number of moles of each reagent...............and calculate $\left[H N {O}_{3}\right]$ to be a bit over $0.2 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

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

$N {a}_{2} C {O}_{3} \left(a q\right) + 2 H N {O}_{3} \left(a q\right) \rightarrow 2 N a N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right) + C {O}_{2} \left(g\right) \uparrow$

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

$\text{Moles of sodium carbonate } =$

$\frac{0.425 \cdot g}{105.99 \cdot g \cdot m o {l}^{-} 1} = 4.01 \times {10}^{-} 3 \cdot m o l$

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:

$\left[H N {O}_{3}\right] = \frac{2 \times 4.01 \times {10}^{-} 3 \cdot m o l}{33.25 \cdot m L} \times {10}^{3} \cdot m L \cdot {L}^{-} 1 \cong$

$0.25 \cdot m o l \cdot {L}^{-} 1$.