What is the concentration of the solution that results from mixing 40.0 $\mathrm{mL}$of 0.200 M $\mathrm{HCl}$with 60.0 $\mathrm{mL}$of 0.100 $\mathrm{M}$$\mathrm{NaOH}$? (You may assume the volumes are additive.)

the concentration of the solution that results from mixing 40.0 $\mathrm{mL}$of 0.200 M $\mathrm{HCl}$with 60.0 $\mathrm{mL}$of 0.100 $\mathrm{M}$$\mathrm{NaOH}$ is :

$\mathrm{HCl}+\mathrm{NaOH}\to \mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} \text{ Moles }\Rightarrow (40\times 0.2)(60\times 0.1) \\ 8 6 \end{gathered}$$

      $\left[\mathrm{NaCl}\right]=\frac{6\times {10}^{-3}}{100\times {10}^{-3}l}\left[\mathrm{HCl}\right]=\frac{2\times {10}^{-3}}{100\times {10}^{-3}l}$

$\mathrm{NaCl}=0.06\mathrm{m}\mathrm{HCl}=0.02\mathrm{m}$