This problem(from ionic equiibrium) states:
We are given $\pu{40 mL}$ of $\pu{0.1 M}$ $\ce{HCl}$ and $\pu{10 mL}$ of $\pu{0.45 M}$ $\ce{NaOH}$ in two separate beakers. These two solutions are mixed. Find the $\mathrm {pH}$ of the resultant solution formed. Consider the concentration of the ions formed by self dissociation of water to be negligible.
My method
Number of moles of $\ce{H+}$ ion in the first solution is $\pu{0.004 mol}$ and the number of moles of $\ce{OH-}$ in the second solution is $\pu{0.0045 mol}$.
Now, after mixing the two solutions, the resulting concentration of $\ce{H+}$ ion becomes $\pu{0.08 M}$ and that of the $\ce{OH-}$ will become $\pu{0.09 M}$. Now these $\ce{H+}$ and $\ce{OH-}$ ions will react with each other such that the equilibrium constant of their reaction becomes $10^{-14}$.
So I formed the equation
$$(0.08-x)(0.09-x) = 10^{-14}$$
From here we can calculate $x$ and subsequently we can get the concentration of $\ce{H+}$ ion in the final solution.
I want to know is this the correct way to solve the problem?
The answer to this problem is given as $12$, which I doubt can be obtained from my way of doing the problem.