Question

You have 75.0 mL of a 2.50 M solution of Na₂CrO₄(aq). You also have 125 mL of a 1.52 M solution of AgNO₃(aq). Calculate the concentration of CrO₄^2– after the two solutions are mixed together.

Answer 1

Molarity is the number of moles of solute present in one litre of solution.

Number of moles of solute

= (M *Volumeof solution in mL)/1000

Number of moles Na₂CrO₄ = 75.0 mL*2.5 M/1000

= 0.1875 moles

Number of moles of AgNO_​​​3 = 125 mL*1.52 M/1000

= 0.19 moles

Total Concentration of chromate ions present in 200 mL of solution = 0.1875*200/1000

= 0.0375M

When 0.19 moles of AgNO3 is mixed with 0.1875 moles of sodium chromate, 0.1875 moles of Ag2CrO4 is formed. 0.0025 moles of chromate ion (CrO₄^2-) is present in free form.

Therefore 0.0025 moles of chromate ions present in 200 mL of solution and is present as free chromate ions.

Concentration of chromate ions = 0.0025*200/1000

= 0.0005 M