# how to calculate the initial concentration of NaOH in a mixture? [closed]

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)

• Right equation. The gist here is that you must assume that the volumes are additive. That isn't strictly true, but since both solutions are fairly dilute the assumption is "good enough" for this problem. – MaxW Mar 12 '19 at 18:05
• @MaxW but how do I find the total concentration of the reaction? I don't think I can just sum up both concentrations. I'm confused. – nutshell_A Mar 12 '19 at 18:10
• ? I said "you must assume that the volumes are additive." So the total volume is 20+20=40 ml – MaxW Mar 12 '19 at 18:56

## 1 Answer

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: $$\frac{\overset{\text{this are your moles of solute}}{\overbrace{\frac{\text{0.0400 mol}}{\text{1 L}}\times{\text{20 mL}}}}}{\underset{\text{this is your final volume}}{\underbrace{\text{40 mL}}}}=\text{0.0400 }\frac{\text{mol}}{\text{L}}\times\frac{1}{2}=\text{0.0200 }\frac{\text{mol}}{\text{L}}$$