Start by writing the balanced chemical equation for the double replacement reaction that takes place between nickel (II) sulfate, NiSO_4NiSO4, and potassium hydroxide, KOHKOH
NiSO_(4(aq)) + color(red)(2)KOH_((aq)) -> Ni(OH)_(2(s)) + K_2SO_(4(aq))NiSO4(aq)+2KOH(aq)→Ni(OH)2(s)+K2SO4(aq)
You can determine the concentration of the K^(+)K+ and SO_4^(2-)SO2−4 ions just be inspection of the complete ionic equation
Ni_((aq))^(2+) + SO_(4(aq))^(2-) + 2K_((aq))^(+) + 2OH_((aq))^(-) -> Ni(OH)_text(2(s]) + 2K_text((aq])^(+) + 2SO_(4(aq))^(2-)Ni2+(aq)+SO2−4(aq)+2K+(aq)+2OH−(aq)→Ni(OH)2(s]+2K+(aq]+2SO2−4(aq)
Notice that both the K^(+)K+ and SO_4^(2-)SO2−4 ions are actually spectator ions; since they can be found on both sides of the reaction, these ions will not participate in the reaction.
As a result, their initial concentration will change solely by dilution, since the volume of the final solution will be equal to the sum of the two added volumes.
Use the molarities of the two initial solutions to determine how many moles of each you add to the mix
C = n/V => n = C * VC=nV⇒n=C⋅V
n_(NiSO_4) = 116 * 18^(-3)"L" * "0.190 M" = "0.02204 moles "nNiSO4=116⋅18−3L⋅0.190 M=0.02204 moles NiSO_4NiSO4
n_(KOH) = 210 * 10^(-3)"L" * "0.240 M" = "0.0504 moles KOH"nKOH=210⋅10−3L⋅0.240 M=0.0504 moles KOH
Since both compounds are soluble in aqueous solution, you'll get
n_(KOH) = n_(K^(+)) = n_(OH^(-)) = "0.0504 moles"nKOH=nK+=nOH−=0.0504 moles
n_(NiSO_4) = n_(Ni^(2+)) + n_(SO_4^(2-)) = "0.02204 moles"nNiSO4=nNi2++nSO2−4=0.02204 moles
The total volume of the solution will be
V_"sol" = V_(KOH) + V_(NiSO_4) = 116 + 210 = "326 mL"Vsol=VKOH+VNiSO4=116+210=326 mL
As a result, you'll get
[K^(+)] = "0.0504 moles"/(326 * 10^(-3)"L") = color(green)("0.155 M")[K+]=0.0504 moles326⋅10−3L=0.155 M
[SO_4^(2-)] = "0.02204 moles"/(326 * 10^(-3)"L") = color(green)("0.0676 M")[SO2−4]=0.02204 moles326⋅10−3L=0.0676 M
SIDE NOTE Alternatively, you can use the dilution calculations equation, C_1V_1 = C_2V_2C1V1=C2V2, to determine the concentrations of these two ions.
Now for the Ni^(2+)Ni2+ cations. The net ionic equation for this reaction is
Ni_((aq))^(2+) + color(red)(2)OH_((aq))^(-) -> Ni(OH)_(2(s))Ni2+(aq)+2OH−(aq)→Ni(OH)2(s)
Notice that you need color(red)(2)2 moles of hydroxide ions for every mole of nickel ions present in solution. This is equivalent to
0.02204cancel("moles "Ni^(2+)) * (color(red)(2)" moles "OH^(-))/(1cancel("mole "Ni^(2+))) = "0.04408 moles " OH^(-)
Since you have 0.0504 moles of hydroxide ions in solution, the Ni^(2+) ions will act as a limiting reagent.
This means that all of the Ni^(2+) cations will be consumed by the reaction and you'll end up with excess hydroxide ions. As a result,
[Ni^(2+)] = color(green)("0 M")
SIDE NOTE If you use the solubility product constant, K_(sp), of nickel (II) hydroxide in order to determine what concentration of nickel cations you'd have in solution, the answer should be similar to
[Ni^(2+)] = 2.84 * 10^(-14)"M"
For all intended purposes, that concentration is zero.
