Question

What concentration of NO−3NO3− results when 897 mL897 mL of 0.497 M NaNO30.497 M NaNO3 is mixed with 813 mL813 mL of 0.341 M Ca(NO3)2?

Answer 1

Step-by-step explanation:

NaNO₃ = Na⁺ + NO₃⁻¹

.497 M .497 M

moles of NO₃⁻¹ = .897 x .497 = .4458 moles

Ca( NO₃)₂ = Ca + 2 NO₃⁻¹

.341 M 2 x .341 M = .682 M

moles of NO₃⁻¹ = .813 x .682 = .5544 moles

Total moles = .4458 moles + .5544 moles

= 1.0002 moles

volume of solution = 897 + 813 = 1710 mL

= 1.710 L

concentration of nitrate ion = 1.0002 / 1.710 M

= .585 M