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The question in my assignment is as follows. You have 5.5 M $\ce{NaOH}$. The volume of $\ce{NaOH}$ is 0.5 L. How many mL of the 5.5 M $\ce{NaOH}$ solution do you need if you would like to make 300 mL of 1.2 M $\ce{NaOH}$? I thought you could use the $C_1V_1= C_2V_2$ formula and rearrange it in there to get the answer, 720 ml. But that doesn't quite feel right. Does anyone have hints on what I'm doing wrong?

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  • $\begingroup$ You may take 65 ml of the solution and top it up with 235 ml of water. $\endgroup$
    – Shub
    Aug 24 at 15:24
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    $\begingroup$ It is very simple algebra of junior high school, just darkened by almighty magic, because it is about chemistry. $\endgroup$
    – Poutnik
    Aug 24 at 17:16
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    $\begingroup$ @Shub No. That's not how it is done. $\endgroup$
    – Loong
    Aug 24 at 17:27
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Take care ! The volumes are not additive! A $5.5$ M solution of $\ce{NaOH}$ has a density is $1.205$ and contains $18.1$% $\ce{NaOH}$ by weight. So let's start from the beginning!

A volume $\pu{V}$ of this concentrated solution has to be taken for the dilution. It contains $\pu{5.5·V}$ moles of $\ce{NaOH}$. The final solution must contain $\pu{1.2 mol L^{-1}*~ 0.3 L~ =~ 0.36~ mol}$ $\ce{NaOH}$. As these two number of moles are equal, the volume $\pu{V}$ of the concentrated solution to be taken for the final dilution is given by : $\pu{5.5 mol L^{-1}*V~ =~ 0.36 mol}$, and $\pu{~ V = \frac {0.36~ mol}{5.5~ mol~ L^{-1}} =~ 0.0655 L~ =~ 65.5 mL}$

To obtain the final solution you should not mix it with $\pu{300 mL - 65.5 mL = 234.5 mL}$ water, as the volumes are not additives. $1 $ liter of the concentrated solution plus $1$ liter water does not make $2$ liters diluted solution. In this case choose a flask containing exactly $\pu{300 mL}$. Transfer $\pu{65.5 mL}$ of the concentrated solution, and then add some water and mix. Then add enough water and mix to get to the exact final volume $\pu{300 mL}$.

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Maurice has already explained it nicely!

You have the right dilution formula, but most students even get this wrong anyway. I substitute the indexes and it helps students remember. You should be able to derive this relationship otherwise it will keep haunting as a mystery formula.

$$C_iV_i= C_fV_f$$

where $C_i$ is the initial concentration of the stock solution

$V_i$ is the initial volume you need from the stock solution

$C_f$ is the final desired concentration of the diluted solution

$V_f$ is the final desired volume of the diluted solution

If you now plug in the numbers you should the answer right, 65.45 mL or 65.5 mL.

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