Question

A 50.0 mL sample of 0.10 M potassium nitrate is mixed with 100 mL of 0.25 M magnesium nitrate. What is the final concentration of nitrate in the solution

Answer 1

The final concentration of nitrate is 0.367 M

Concentration of a substance

The concentration of a substance is the amount of substance present in a given volume of a solution of that substance.

Concentration = number of moles/ volume

Equation for the dissociation of potassium nitrate

KNO3 ---> K+ + NO3-

1 mole of KNO3 produces 1 mole of nitrate

moles of nitrate in a 0.1 M KNO3 solution

• moles = concentration × volume

volume of KNO3 = 50.0 mL= 0.05 L

moles of nitrate = 0.1 × 0.05

moles of nitrate in KNO3 = 0.005

Equation for the dissociation magnesium nitrate

• Mg(NO3)2 ---> Mg^2+ + 2 NO3-

1 mole of Mg(NO3)2 produces 2 moles of nitrate ion

moles of Mg(NO3)2 = concentration × volume

volume of solution = 100 mL= 0.1 L

moles of Mg(NO3)2 = 0.1 × 0.25 = 0.025

Moles of nitrate = 0.025 × = 0.05

Total moles of nitrate = 0.05 + 0.005 = 0.055 moles

Total volume of solution = 0.1 + 0.05 = 0.15 L

Concentration of nitrate = 0.055/0.15

Concentration of nitrate = 0.367 M

Therefore, the final concentration of nitrate is 0.367 M

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