calculate the final concentration of this equation
[&hellip;] he should had used the moles of potassium iodide since it is the limiting reactant.



Did any reaction take place upon mixing solutions of $\ce{Pb(NO3)2}$ and $\ce{KI}$?
If so, did the reaction change the number of moles of $\ce{NO3-}$ in solution by precipitation, oxidation, reduction, etc.?
Would the result of the calculation be different if just 300 mL of water were added to the $\ce{Pb(NO3)2}$ solution?


Update I

Now for the rocket science (solubilities at 20 °C):

\begin{array}{cr}
\mathrm{compound} &amp; \mathrm{solubility\ [g\cdot L^{-1}]} \\ \hline
\ce{Pb(NO3)2} &amp; 522\\ 
\ce{KI} &amp; 1430\\
\ce{KNO3} &amp; 316\\
\ce{PbI2} &amp; 0.76\\
\end{array}

The rather low solubility of $\ce{PbI2}$ suggests that on combination of the two solutions, the following may happen:

$$\ce{Pb^{2+} + 2 NO3- + 2 K+ + 2 I- -&gt; 2 K+ + 2 NO3- + PbI2 v}$$

So again: 


Does this in any way change the absolute amount of  $\ce{NO3-}$ in solution?
What does change for the concentration of $\ce{NO3-}$ when 300 mL of another solution is added to 260ml of 1.2 M $\ce{Pb(NO3)2}$ solution?


Update II


Initial $\ce{Pb(NO3)2}$ solution



   260 mL of a 1.2M solution of $\ce{Pb(NO3)2}$ contain $1.2\ \mathrm{\frac{mol}{L}}\ \cdot\ 2\cdot\ 0.26\ \mathrm{L} = 0.624\ \mathrm{mol}\ \ce{NO3-}$ 



After mixing



   Addition of 300 mL of the $\ce{KI}$ solution results in the precipitation of $\ce{PbI2}$, but does not change the absolute amount of  $\ce{NO3-}$ in solution. However, it changes the total volume and hence the concentration, which calculates to $\frac{0.624\ \mathrm{mol}}{(0.26 + 0.3) \mathrm{L}} = 1.11 \ M$