10 mL of a solution of $\mathit{NaOH}$ is found to be completely neutralized by 8 mL of a given solution of $\mathit{HCl}$. If we take 20 mL of the same solution of $\mathit{NaOH}$, what will be the amount of $\mathit{HCl}$ solution required to neutralize it?

20 mL of $\mathit{NaOH}$ solution will be neutralized by 16 mL of $\mathit{HCl}$ solution.

Given 10 mL of $\mathit{NaOH}$ solution is neutralized by 8 mL of the $\mathit{HCl}$solution. 

Since, 

8 mL of $\mathit{HCl}$ is needed for 10 mL of $\mathit{NaOH}$.

Let X be the volume of $\mathit{HCl}$ needed to dissolve 20 milliliters of $\mathit{NaOH}$.
⇒$\mathit{X }=\frac{ (20 \times 8)}{10}$
⇒$\mathit{X }= 16\mathit{ mL}$

So, 20 mL of $\mathit{NaOH}$ solution will need 16 mL of $\mathit{HCl}$ solution to neutralize.