0
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$$c_1\times V_1=c_2\times V_2$$

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

$$\begin{align} c_1\times V_1&=c_2\times V_2\\[6pt] V_1 & =\frac{c_2\times V_2}{c_1}\\[6pt] & =\frac{\pu{3.50mol L^{-1}}\times \pu{75.0mL}}{\pu{12.0mol L^{-1}}}\\[6pt] & =\pu{21.875 mL}\\[6pt] & \approx \pu{21.9mL} \end{align}$$

|improve this answer|||||
$\endgroup$
1
$\begingroup$

"$\pu{12 M }\ce{ HCl}$" 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.

|improve this answer|||||
$\endgroup$
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.