Dilution calculations $\pu{5 mL}$ of $\pu{0.012 M}\,\ce{Pb(NO3)2}$ is mixed with $\pu{5 mL}$ of $\pu{0.030 M}\,\ce{KI}$. $\ce{PbI2}$ forms. Calculate the diluted concentrations of $\ce{Pb^{2+}}$ and $\ce{I-}$ ions.


I couldn't figure out whether it is an equilibrium question, or something else entirely.

I have currently tried using the moles of the limiting reactant (Pb(NO3)2) to obtain the moles of PbI2 formed and using the molar ratio to get concentration of each ion. However I have no idea whether I have done the correct calculation for ion concentration. The values I am getting are 0.003M Pb2+ and 0.012M I-

update: I have looked at the questions that follow and am able to say for certain this is not an equilibrium question.
I haven't checked your numbers but you're assuming the $\ce{PbI2}$ to be perfectly insoluble. That's a reasonable assumption but in reality its solubility product is $K_{sp}=4.41\times 10^{-9}$ (Wikipedia), so it does have some solubility.

To take this into account it becomes an equilibrium problem. Assume the initial concentrations (just after mixing)  to be $\ce{C(Pb^2+)}$ and $\ce{C(I^-)}$.

After precipitation the lead concentration has been reduced by$X$ and the iodide concentration by $2X$.

With the solubility product after precipation:

$$K_{sp}=[\ce{Pb^2+}]\times [\ce{I^-}]^2$$

So we have:

$$K_{sp}=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

Note that this is a third degree polynomial equation in $X$. Solve for $X$ to find the real equilibrium concentrations $[\ce{Pb^2+}]$ and $[\ce{I^-}]$.

We can also use the equation for the case where $K_{sp}=0$. Then:

$$0=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

First case: $\ce{C(Pb^2+)}-X=0 \implies \ce{C(Pb^2+)}=X \implies [\ce{Pb^2+}]=0$

Also: $[\ce{I^-}]=\ce{C(I^-)}-2\times\ce{C(Pb^2+)}$

In the second case we assume $\Big(\ce{C(I^-)}-2X\Big)^2=0$ and proceed as above. 

Which case applies depends on which reagent is the limiting one.