Question

A solution of 116 mL of 0.170 M KOH is mixed with a solution of 300 mL of 0.200 M NiSO4. What is the concentration of Ni2+ that remains in solution? What is the concentration of SO2−4 that remains in solution? What is the concentration of K+ that remains in solution?

Answer 1

Reaction Involved : 2KOH + NiSO₄ -------> Ni(OH)₂ ppt + K₂SO₄

Moles of KOH = 0.116L *0.170 M = 0.01972 ( 1L = 1000 mL)

Moles of NiSO₄ = 0.3 L* 0.200 M = 0.006

From the balanced reaction shown above it is found that the 1 mole of NiSO₄ is react with 2 moles of KOH

Therefore,0.01972 moles of KOH *(1/2) = 0.0098 moles of NiSO₄ consumed and 0.0038 moles (0.0098-0.006) of NiSO₄ unreacted and the total volume = 0.116L+0.3L =0.416 L

The Concentration of Ni²⁺ that remains in the solution = 0.0038/0.416 = 0.0091M

The Concentration of SO_4^2- that remains in the solution = 0.006/0.416 = 0.0144M

The Concentration of K⁺ that remains in the solution = 0.01972/0.416 = 0.04740M