how to calculate the initial concentration of NaOH in a mixture? I have doubts about how to find the equation for the calculation of the initial concentration of $\ce{NaOH}$. I have the following information:
 in the kinetic study of a reaction,

We measure $\ce{20.00 ml}$ of $\ce{CH_3COOC_2H_5}$ solution of 0.0400 $\ce{mol/L}$ and also $\ce{20.00 ml}$ of $\ce{NaOH}$ solution of 0.0400 $\ce{mol/L}$

We mix both substances in one Erlenmeyer.

The question is:  find the equation to calculate the concentration of $\ce{NaOH}$ at $\ce{t=0}$.

It is a bit weird to me because a concentration of the $\ce{NaOH}$ solution is already given.

The best (and the only) that I can think is this:
if the concentrations of the reactives are equal, then:
$$\ce{C_iV_i = C_fV_f}$$

then we have,
$$\ce{C_i=C_fV_f/V_i}$$

$\ce{C_i}$ is the initial concentration of $\ce{NaOH}$, $\ce{V_i}$ is the initial volume of $\ce{NaOH}$, $\ce{C_f}$ is the final concentration of the whole mixture ( which I don't know how to find), and $\ce{V_f}$ is the final volume of the whole mixture

Probably this is wrong but I don't have any other ideas.
 any clues?

Thanks.

(Note: apologies for my poor English)
As @MaxW pointed out, if you assume that the volumes are additive implies that your total volume is 40 mL. Now, remember the definition of molar concentration:
$$C=\frac{\text{moles solute}}{\text{total volume}}=\frac{n}{V}$$

That means that the concentration and the volume are inversely proportional, thus if initially, you have a $\text{0.0400 } \frac{\text{mol}}{\text{L}}$ in 20 mL and then you increase the volume twice, then your concentration decrease to the half of initial value. In numbers:
$$\frac{\overset{\text{this are your moles of solute}}{\overbrace{\frac{\text{0.0400 mol}}{\text{1 L}}\times{\text{20 mL}}}}}{\underset{\text{this is your final volume}}{\underbrace{\text{40 mL}}}}=\text{0.0400 }\frac{\text{mol}}{\text{L}}\times\frac{1}{2}=\text{0.0200 }\frac{\text{mol}}{\text{L}} $$