Question

`1.00 *10^2` `"mL"` of a `"12.4-M"` `"HCl"` solution is added to water to bring the total volume of the solution to `"0.820 L"`. What is the concentration of this new solution?

Answer 1

`"1.51 M"`

Explanation:

You know that you're diluting `1.00 * 10^2 quad "mL"` of a `"12-4M"` hydrochloric acid solution by adding it to enough water to get the total volume to

`0.820 color(red)(cancel(color(black)("L"))) * (10^3 quad "mL")/(1color(red)(cancel(color(black)("L")))) = "820. mL"`

Now, when you're diluting a solution, you must keep in mind that the ratio that exists between the volume of the diluted solution and the volume of the concentrated solution is equal to the dilution factor, `"DF"`.

`"DF" = V_"diluted"/V_"stock"`

Moreover, the dilution factor is also equal to the ratio that exists between the concentration of the concentrated solution and the concentration of the diluted solution.

`"DF" = c_"stock"/c_"diluted"`

In your case, the dilution factor is equal to

`"DF" = (820. color(red)(cancel(color(black)("mL"))))/(1.00 * 10^2color(red)(cancel(color(black)("L")))) = color(blue)(8.20)`

You can thus say that the concentration of the initial solution, i.e. the concentrated solution, was `color(blue)(8.20)` times higher than the concentration of the diluted solution, and so

`c_"diluted" = c_"stock"/color(blue)(8.20)`

`c_"diluted" = "12.4 M"/color(blue)(8.20) = color(darkgreen)(ul(color(black)("1.51 M")))`

The answer is rounded to three sig figs.

So, if you start with `1.00 * 10^3 quad "mL"` of a `"12.4-M"` hydrochloric acid solution and add it to enough water to get its total volume of `"0.820 L"`, you will end up with `"0.820 L"` of a `"1.51-M"` hydrochloric acid solution.

Don't forget that when diluting strong acids, you must always add the acid to the water and not the water to the acid!