# Question 192a2

Apr 23, 2015

Start by writing the balanced chemical equation for the double replacement reaction that takes place between nickel (II) sulfate, $N i S {O}_{4}$, and potassium hydroxide, $K O H$

$N i S {O}_{4 \left(a q\right)} + \textcolor{red}{2} K O {H}_{\left(a q\right)} \to N i {\left(O H\right)}_{2 \left(s\right)} + {K}_{2} S {O}_{4 \left(a q\right)}$

You can determine the concentration of the ${K}^{+}$ and $S {O}_{4}^{2 -}$ ions just be inspection of the complete ionic equation

$N {i}_{\left(a q\right)}^{2 +} + S {O}_{4 \left(a q\right)}^{2 -} + 2 {K}_{\left(a q\right)}^{+} + 2 O {H}_{\left(a q\right)}^{-} \to N i {\left(O H\right)}_{\textrm{2 \left(s\right]}} + 2 {K}_{\textrm{\left(a q\right]}}^{+} + 2 S {O}_{4 \left(a q\right)}^{2 -}$

Notice that both the ${K}^{+}$ and $S {O}_{4}^{2 -}$ 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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N i S {O}_{4}} = 116 \cdot {18}^{- 3} \text{L" * "0.190 M" = "0.02204 moles }$ $N i S {O}_{4}$

${n}_{K O H} = 210 \cdot {10}^{- 3} \text{L" * "0.240 M" = "0.0504 moles KOH}$

Since both compounds are soluble in aqueous solution, you'll get

${n}_{K O H} = {n}_{{K}^{+}} = {n}_{O {H}^{-}} = \text{0.0504 moles}$

${n}_{N i S {O}_{4}} = {n}_{N {i}^{2 +}} + {n}_{S {O}_{4}^{2 -}} = \text{0.02204 moles}$

The total volume of the solution will be

${V}_{\text{sol" = V_(KOH) + V_(NiSO_4) = 116 + 210 = "326 mL}}$

As a result, you'll get

[K^(+)] = "0.0504 moles"/(326 * 10^(-3)"L") = color(green)("0.155 M")

[SO_4^(2-)] = "0.02204 moles"/(326 * 10^(-3)"L") = color(green)("0.0676 M")

SIDE NOTE Alternatively, you can use the dilution calculations equation, ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, to determine the concentrations of these two ions.

Now for the $N {i}^{2 +}$ cations. The net ionic equation for this reaction is

$N {i}_{\left(a q\right)}^{2 +} + \textcolor{red}{2} O {H}_{\left(a q\right)}^{-} \to N i {\left(O H\right)}_{2 \left(s\right)}$

Notice that you need $\textcolor{red}{2}$ 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 "# $O {H}^{-}$

Since you have 0.0504 moles of hydroxide ions in solution, the $N {i}^{2 +}$ ions will act as a limiting reagent.

This means that all of the $N {i}^{2 +}$ cations will be consumed by the reaction and you'll end up with excess hydroxide ions. As a result,

$\left[N {i}^{2 +}\right] = \textcolor{g r e e n}{\text{0 M}}$

SIDE NOTE If you use the solubility product constant, ${K}_{s p}$, of nickel (II) hydroxide in order to determine what concentration of nickel cations you'd have in solution, the answer should be similar to

$\left[N {i}^{2 +}\right] = 2.84 \cdot {10}^{- 14} \text{M}$

For all intended purposes, that concentration is zero.