What is the concentration of \([ \text{NO}_3^- ]\) in a solution made up of 37.9 mL of 0.197M \(\text{NaNO}_3\) and 42.5 mL of 0.378M \(\text{Bi(NO}_3\text{)}_3\)?

To find the concentration of NO₃⁻ in the solution, we can use the concept of dilution. When two solutions are mixed, the moles of solute before and after mixing remain the same. Therefore, the initial moles of NO₃⁻ in the solution should equal the final moles of NO₃⁻ in the solution.

Given:

- Volume of NaNO₃ solution (V₁) = 37.9 mL

- Concentration of NaNO₃ solution (C₁) = 0.197 M

- Volume of Bi(NO₃)₃ solution (V₂) = 42.5 mL

- Concentration of Bi(NO₃)₃ solution (C₂) = 0.378 M

First, we calculate the moles of NO₃⁻ from each solution:

Moles of NO₃⁻ from NaNO₃ solution = V₁ [tex]\times[/tex] C₁ = 37.9 [tex]\times[/tex] 10³ L [tex]} \times[/tex] 0.197 M

Moles of NO₃⁻ from Bi(NO₃)₃ solution = V₂ [tex]\times[/tex] C₂ = 42.5 [tex]\times[/tex] 10⁻³ L[tex]\times[/tex] 0.378 M

The total moles of NO₃⁻ in the solution = Moles from NaNO₃ solution + Moles from Bi(NO₃)₃ solution

Finally, we calculate the concentration of NO₃⁻ in the solution:

Concentration of NO₃⁻ = Total moles of NO₃⁻ / Total volume of the solution

Substitute the values and calculate the concentration of NO₃⁻.

The concentration of NO₃⁻ in the solution is approximately 0.323 M.

In conclusion, the concentration of NO₃⁻ in the solution is determined by considering the initial moles of NO₃⁻ from both solutions and dividing by the total volume of the solution.

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