Question

In the laboratory, a student combines 25.1 mL of a 0.296 M sodium nitrate solution with 13.3 mL of a 0.402 M silver nitrate solution. What is the final concentration of nitrate anion?

a) 0.298 M
b) 0.304 M
c) 0.398 M
d) 0.402 M

Answer 1

The final concentration of nitrate anion after mixing sodium nitrate and silver nitrate solutions is approximately 0.332883333 M, which doesn't match precisely with any given options.

Step-by-step explanation:

To calculate the final concentration of nitrate anion after mixing two solutions, we must first determine the total moles of nitrate in both solutions and then divide by the total final volume of the mixture.

For sodium nitrate (NaNO3), we have:

• Volume = 25.1 mL = 0.0251 L
• Molarity (M) = 0.296 mol/L
• Moles of nitrate = 0.0251 L × 0.296 mol/L = 0.0074296 mol

For silver nitrate (AgNO3), we have:

• Volume = 13.3 mL = 0.0133 L
• Molarity (M) = 0.402 mol/L
• Moles of nitrate = 0.0133 L × 0.402 mol/L = 0.0053466 mol

Total moles of nitrate = 0.0074296 mol + 0.0053466 mol = 0.0127762 mol

Total volume of solution = 25.1 mL + 13.3 mL = 38.4 mL = 0.0384 L

The final concentration of nitrate anion is:

Final concentration = Total moles of nitrate / Total volume

Final concentration = 0.0127762 mol / 0.0384 L

Final concentration = 0.332883333 M

This final concentration doesn't match exactly with any of the options provided. The closest answer, taking into account potential rounding differences during the calculation, would be (c) 0.398 M, though this still isn't an exact match.