Ch 202 problem 4.87

4.87 A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiSO4.

(a) Write the balanced chemical equation for the reaction that occurs.

nice review!!! gotta make sure you still have mad skillz in writing reactions. This is just a D-D reaction from last term.

2KOH(aq) + NiSO4(aq) -- > Ni(OH)2(s) + K2SO4(aq)

note that the solubility table (table 4.1) was used to deduce that nickel (II) hydroxide is insoluble.

(b) What precipitate forms? from part (a), Ni(OH)2.

(c) What is the limiting reactant?

YES! More mad skillz review:) Remeber how to do these? We figure out how many mol of each reactant we have (use molarity x volume), and then we identify the limiting reactant by converting mol reactant A --> mol B, and vice versa.....

starting mol KOH

starting mol NiSO4

now convert mol KOH --> mol NiSO4

need 0.0100 mol NiSO4 to consume 0.020 mol KOH

convert mol NiSO4 -> mol KOH

need 0.0600 mol KOH to consume 0.030 mol NiSO4. We don't have enough KOH, so KOH is the limiting reactant.

now to find out how many g of precipitate form we convert mol limiting reactant to g product...

molar mass of the product (Ni(OH)2)....

go for it....

theoretical yield of Ni(OH)2!

(d) What is the concentration of each ion that remains in solution?

Alrighty. In part (a) we write the balanced equation for the reaction which occurred...

2KOH(aq) + NiSO4(aq) -- > Ni(OH)2(s) + K2SO4(aq)

We can write this as a complete ionic equation using what we know about soluble strong electrolytes...

2K+(aq) + 2OH-(aq) + Ni2+(aq) + SO42-(aq) --> Ni(OH)2(s) + 2K+(aq) + SO42-(aq)

So, in net ionic form, we would have

2OH-(aq) + Ni2+(aq) --> Ni(OH)2(s)

The point here is this: K+(aq) and SO42-are spectator ions in this reaction - the number of mol of these species doesn't change. Their concentrations change only by dilution. So, we started with 0.0200 mol KOH = 0.0200 mol K+, and the final volume of the solution is 300 mL,

concetration of K+ remaining

similarly, we started with 0.03 mol NiSO4 = 0.03 mol SO42- and the final volume is 300 mL:

concentration of SO42-

Now, the OH- was found to be the limiting reactant (we did this in part (c), so there is no OH- left in solution.

Since NiSO4 was found to be the XS reactant, we should be able to calculate how much is left in the usual way. We started with 0.03 mol NiSO4, and above we found that we needed 0.01 mol NiSO4 to consume all of the KOH.

mol NiSO4 left

this is now in a volume of 300 mL, so the remaining concentration of Ni2+ is

DAC 8/4/14 Nice problem!