What volume of $\pu{12.0 M }\ce{HCl}$ is required to make $\pu{75.0 mL}$ of $\pu{3.50 M }\ce{HCl}$?

I didn't know how to solve it at first until I multiplied $\pu{75mL}$ by $\pu{3.50 M}$ and got $\pu{262.5 mM}$ and then divided that by $\pu{12.0M}$ to get the answer. I was just wondering if there was an equation that I can use for these kinds of questions? Or is it just simple math? Ex. $\pu{75.0mL} \times \pu{3.50 M }\ce{HCl}$.


The equation you are looking for is the definition of concentration

$$c=\frac nV$$

$c$ is concentration,
$n$ is amount of substance, and
$V$ is volume.

You can rearrange this equation to solve for concentration, amount of substance, or volume as required.

Furthermore, you know that the amount of solute does not change when you dilute a solution, i.e.


and thus using the above-mentioned equation

$$c_1\cdot V_1=c_2\cdot V_2$$

You can rearrange this equation to solve the problem that is given in the question:

$$\begin{align} c_1\cdot V_1&=c_2\cdot V_2\\[6pt] V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt] &=\frac{3.50\ \mathrm{mol\ l^{-1}}\times75.0\ \mathrm{ml}}{12.0\ \mathrm{mol\ l^{-1}}}\\[6pt] &=21.875\ \mathrm{ml}\\[6pt] &\approx21.9\ \mathrm{ml} \end{align}$$


"$\pu{12 M }\ce{ HCl}$" means that there are $\pu{12 mol}$ of $\ce{HCl}$ per liter of solution (or 12 mmol per milliliter).

You are looking for a volume of $\pu{12 M }\ce{ HCl}$ that contains ($\pu{3.5\times 75 = 262.5 mmol}$ ) of $\ce{HCl}$.

Since you need $\pu{262.5 mmol}$, and the $\pu{12 M }$ solution has $\pu{12 mmol}$ per $\pu{mL}$, then you need $\pu{262.5mmol}\div \pu{12M} = \pu{21.875 ml}$ of $\pu{12 M }\ce{ HCl}$.

This is to explain why it worked, but I suggest you to learn what the equations mean, rather than just learning "how to do it": once you understand what you are doing, it's not hard.

  • 2
    $\begingroup$ I'd add that using dimensional analysis is a good way to keep things straight until you're comfortably just grinding the numbers. $\endgroup$ – MaxW Sep 9 '16 at 23:52

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