Certainly! Let's work through this problem step-by-step to find the molarity of the nitric acid solution.
### Step 1: Calculate the moles of Sodium Carbonate ([tex]$Na_2CO_3$[/tex])
Given:
- Mass of [tex]$Na_2CO_3$[/tex] = 0.500 g
- Molar mass of [tex]$Na_2CO_3$[/tex] = 105.99 g/mol
The number of moles of [tex]$Na_2CO_3$[/tex] is calculated using the formula:
[tex]\[ \text{Moles of } Na_2CO_3 = \frac{\text{Mass of } Na_2CO_3}{\text{Molar mass of } Na_2CO_3} \][/tex]
So:
[tex]\[ \text{Moles of } Na_2CO_3 = \frac{0.500 \text{ g}}{105.99 \text{ g/mol}} = 0.004717426 \text{ moles} \][/tex]
### Step 2: Calculate the moles of Nitric Acid ([tex]$HNO_3$[/tex])
From the balanced chemical equation:
[tex]\[ 2 HNO_3(aq) + Na_2 CO_3(s) \rightarrow 2 NaNO_3(aq) + H_2 O(l) + CO_2(g) \][/tex]
We see that 2 moles of HNO_3 react with 1 mole of [tex]$Na_2CO_3$[/tex].
Therefore, the moles of HNO_3 required are:
[tex]\[ \text{Moles of } HNO_3 = 2 \times \text{Moles of } Na_2CO_3 \][/tex]
So:
[tex]\[ \text{Moles of } HNO_3 = 2 \times 0.004717426 \text{ moles} = 0.009434852 \text{ moles} \][/tex]
### Step 3: Convert the volume of Nitric Acid solution to liters
Given:
- Volume of [tex]$HNO_3$[/tex] solution = 38.50 mL
To convert this volume to liters, we use the relationship:
[tex]\[ 1 \text{ mL} = 0.001 \text{ liters} \][/tex]
So:
[tex]\[ \text{Volume of } HNO_3 \text{ in liters} = 38.50 \text{ mL} \times 0.001 = 0.03850 \text{ liters} \][/tex]
### Step 4: Calculate the molarity of the Nitric Acid solution
Molarity (M) is defined as the number of moles of solute (in this case, HNO_3) per liter of solution.
[tex]\[ \text{Molarity} = \frac{\text{Moles of } HNO_3}{\text{Volume of solution in liters}} \][/tex]
So:
[tex]\[ \text{Molarity of } HNO_3 = \frac{0.009434852 \text{ moles}}{0.03850 \text{ liters}} = 0.245061 \text{ M} \][/tex]
### Summary
- Moles of [tex]$Na_2CO_3$[/tex] = 0.004717426 moles
- Moles of [tex]$HNO_3$[/tex] = 0.009434852 moles
- Molarity of [tex]$HNO_3$[/tex] = 0.245061 M
Therefore, the molarity of the nitric acid (HNO_3) solution is approximately 0.245 M.
