# A standard nitric acid solution is prepared using 0.425 g of sodium carbonate, Na_2CO_3. What is the molarity of the acid if 33.25 mL are required to reach a permanent endpoint?

Aug 23, 2016

$\textsf{0.241 \textcolor{w h i t e}{x} \text{mol/l}}$

#### Explanation:

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

Find the no. moles of $\textsf{N {a}_{2} C {O}_{3}}$:

$\textsf{{n}_{N {a}_{2} C {O}_{3}} = \frac{m}{M} _ r = \frac{0.425}{105.99} = 0.004}$

From the equation you can see that the no. moles of $\textsf{H N {O}_{3}}$ must be twice this amount.

$\therefore$$\textsf{{n}_{H N {O}_{3}} = 0.004 \times 2 = 0.008}$

$\textsf{c = \frac{n}{v} = \frac{0.008}{\frac{33.75}{1000}} = 0.241 \textcolor{w h i t e}{x} \text{mol/l}}$