Question

What volume of 3.0 M `HNO_3 (aq)` must be diluted to make 4.0 L of 0.20 M `HNO_3(aq)`?

Answer 1

`"270 mL"`

Explanation:

The key here is the ratio that exists between the initial concentration of the solution, i.e. the concentration of the concentrated solution, and the final concentration of the solution, i.e. the concentration of the diluted solution.

This ratio

`color(blue)(|bar(ul(color(white)(a/a)"DF" = c_"concentrated"/c_"diluted"color(white)(a/a)|)))`

is called the dilution factor and it tells you how concentrated the initial solution was compared with the diluted solution. In this case, you have

`"DF" = (3.0 color(red)(cancel(color(black)("M"))))/(0.20color(red)(cancel(color(black)("M")))) = 15`

This means that the concentrated solution was diluted by a factor of `15` to get the diluted solution. As you know, the underlying principle of a dilution is that the number of moles of solute remains constant.

You can thus decrease the concentration of a given solution by increasing its volume while keeping the number of moles of solute unchanged.

Now, the only way to have a diluted solution that is `15` times less concentrated than the concentrated solution is for it to have a volume that is `15` times bigger than the volume of the concentrated solution.

The dilution factor can thus be expressed in terms of the volume of the concentrated solution, `V_"concentrated"`, and the volume of the diluted solution, `V_"diluted"`, as

`color(blue)(|bar(ul(color(white)(a/a)"DF" = V_"diluted"/V_"concentrated"color(white)(a/a)|)))`

You will have

`V_"concentrated" = V_"diluted"/"DF"`

Plug in your values to find

`V_"concentrated" = "4.0 L"/15 = "0.26667 L"`

Rounded to two sig figs and expressed in milliliters, the answer will be

`V_"concentrated" = color(green)(|bar(ul(color(white)(a/a)color(black)("270 mL")color(white)(a/a)|)))`

So, to make `"4.0 L"` of `"0.20 M"` nitric acid solution, start with `"270 mL"` of `"3.0 M"` nitric acid solution and add enough water to get the total volume of the solution to `"4.0 L"`.