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?

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

Explanation:

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