If 10.0 mL of 0.350 M aqueous NaNO3 is mixed with 50.0 mL of 0.170 M aqueous KNO3, what is the molarity of nitrate ion?

A) 0.185 M

B) 0.200 M

C) 0.260 M

D) 0.520 M

0.200 mol/L

Explanation:

1. Moles of nitrate in first solution

Moles = volume * concentration

n = 0.0100 L * 0.350 mol/L = 0.003 500 mol

2. Moles of nitrate in second solution

n = 0.0500 L * 0.170 mol/L = 0.008 500 mol

3. Total moles of nitrate

n = 0.003 500 + 0.008 500 = 0.012 000 mol

4. Total volume of solution

V = 10.0 + 50.0 = 60.0 mL = 0.0600 L

5. New concentration of nitrate

c = moles/litres

c = 0.012 000 mol/0.0600 L = 0.200 mol/L

The molar concentration of nitrate ion is 0.200 mol/L.