Question

What would the concentration of nitrate be (in M) in a solution prepared by adding 30.0 mL of 0.050 M potassium nitrate with 40.0 mL of a 0.075 M sodium nitrate solution into a volumetric flask that is 250 mL and then filling it up to the 250 mL mark with water and mixing it well?

Answer 1

Potassium Nitrate = KNO3

Sodium Nitrate = NaNO3

Moles = Molarity(M) x Volume(in L)

moles of NO3- = moles of KNO3 , since KNO3 => K+ + NO3-

Simiraly, moles of NO3- = moles of NaNO3 , since NaNO3 => Na+ + NO3-

Total moles of NO3- = moles of KNO3 + moles of NaNO3

moles of KNO3 = 0.05M x (30/1000)L = 1.5 millimoles

moles of NaNO3 = 0.075M x (40/1000)L = 3 millimoles

Total moles of NO3- = 4.5 millimoles

Since the flask is filled up to 250mL mark, the total volume of solution is = 250 mL

The final concentration of NO3- = total moles of NO3- / Total volume of solution

= 4.5x10-3 moles / 250x10-3 L = 0.018M