What remains constant before and after the addition of water is the number of moles in the beaker.
So, let the concentration be c₁ and c₂ (before and after), the number of moles be n and the volume before be V₁ and after be V₂.
The equation for concentration is number of moles divided by the volume, so we have, for before the addition:
[tex]c_1=\frac{n}{V_1}\leftrightarrow n=c_1V_1[/tex]And the equation for after:
[tex]c_2=\frac{n}{V_2}\leftrightarrow n=c_2V_2[/tex]Notice that we didn't put subscripted number on n because this won't change.
Because n doesn't change, we can say:
[tex]\begin{gathered} n=n \\ c_1V_1=c_2V_2 \\ c_2=\frac{c_1V_1}{V_2}_{}_{} \end{gathered}[/tex]We know c₁ is 0.14 mol/L and V₁ is 150 mL.
The volume after the addition will be the volume before plus the amount we added, so:
[tex]V_2=150+350=500[/tex]Thus, the final concentration will be:
[tex]c_2=\frac{0.14mol/L\cdot150\cancel{mL}}{500\cancel{mL}}=0.042mol/L[/tex]The final concentration will be 0.042 mol/L.
