Answered

Find expert answers and community-driven knowledge on IDNLearn.com. Our platform provides prompt, accurate answers from experts ready to assist you with any question you may have.

A nitric acid solution is standardized using 0.500 g of sodium carbonate, [tex]Na_2CO_3[/tex]. Find the molarity of the nitric acid if 38.50 mL are required to reach a permanent endpoint.

[tex]\[ 2 HNO_3(aq) + Na_2CO_3(s) \rightarrow 2 NaNO_3(aq) + H_2O(l) + CO_2(g) \][/tex]

Sagot :

GinnyAnsweridn
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.
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.