A solution of 114mL of 0.150 M KOH is mixed with a solution of 240mL of 0.250 M NiSO4What is the concentration of Ni2+ that rema


A solution of 114mL of 0.150 MKOH is mixed with a solution of 240mL of 0.250 M Niso4. What is the concentration of Ni2+ that remains in solution. 

Assuming the reaction,  2 KOH   +  NiSO4  --->  Ni(OH)2   +   K2SO4

We need to find which of the reactants is in excess.

                 moles of KOH  =  .114 L    x   .150 mol/L   =   .0171 noles of KOH

                   moles of NiSO4   =   .240L   x  .250 mol/L   =  .060 moles of NiSO4

That means that the KOH is the limiting reagent since it will be used up after reacting with only a fraction of the Ni+2 ions that are present.  Based on the chemical equation, we can assume that the will remove .0085 moles of Ni+2 ion.  


Since we started with .060 moles of Ni+2 and we removed .00085 moles, we can determine the amount not precipitated by subtraction:

                      moles of unreacted Ni+2  =     .060   -   .0085  =   .0515 moles

To find the concentration of the NI+2 we need to find the molarity:

                          molarity  =  moles of Ni+2  /   total volume of solution

                                            =   .0515 mol  /  .354 L  =  .15M