How to Find Volume from Moles?

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. $$\pu{75.0mL} \times \pu{3.50 M }\ce{HCl}$$.

2 Answers

The equation you are looking for is the definition of concentration

$$c=\frac nV$$

where
$$c$$ 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.

$$n_1=n_2$$

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.

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