Working out pH of solution when excess NaOH is added In a pH titration, $30\text{cm}^3$ of $\ce{NaOH}$ is added to $\pu{20 cm3}$ $\ce{CH3COOH}$. The concentrations of both solutions was $\pu{0.5 mol dm-3}$. $K_\mathrm{a} = \pu{1.7 x 10^{-5}}$ for $\ce{CH3COOH}$. Calculate the pH of the solution after all the $\ce{NaOH}$ is added.


Clearly, $\pu{10 cm3}~\ce{NaOH}$ is in excess. So moles $\ce{NaOH}$ in excess $= \pu{0.005 mol}$. Then in the mark scheme, it says $$\left[\frac{\ce{NaOH}}{\ce{OH^-}}\right]=0.005 \times \frac{1000}{50} = \pu{0.1 mol dm-3}$$ and I don't understand where this part has come from.
The excess NaOH is, indeed 0.005 moles. The total final volume is 30 ml + 20 ml = 50 ml = 0.05 L. Thus, the final concentration is 0.005 moles/0.05 L = 0.1 M