Question

Question #f9869

Answer 1

Here's what I got.

Explanation:

The idea here is that you need to use the molarity of the three solutions to write two equations that establish a relationship between the number of moles of hydrochloric acid and the volumes of the solutions.

As you know, molarity is defined as the number of moles of solute, which in your case is hydrochloric acid, you get per liter of solution.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid it must contain

`color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))`

You will end up with

`n_(HCl) = "0.2 mol" color(red)(cancel(color(black)("L"^(-1)))) * 2 color(red)(cancel(color(black)("L"))) = "0.40 moles HCl"`

Now, let's assume that you must mix `xcolor(white)(a)"L"` of the `"0.5 M"` solution, i.e. of solution `"A"`, and `y color(white)(a)"L"` of the `"0.1 M"` solution, i.e. of solution `"B"`.

The number of moles of hydrochloric acid coming from solution `"A"` will be

`x color(red)(cancel(color(black)("L"))) * "0.5 moles HCl"/(1color(red)(cancel(color(black)("L")))) = (0.5x)color(white)(a)"moles HCl"`

The number of moles of hydrochloric acid coming from solution `"B"` will be

`y color(red)(cancel(color(black)("L"))) * "0.1 moles HCl"/(1color(red)(cancel(color(black)("L")))) = (0.1y)color(white)(a)"moles HCl"`

Since you know that the target solution must contain `0.40` moles of hydrochloric acid, you an say that

`0.5x + 0.1y = 0.40" " " "color(orange)((1))`

The target solution has a total volume of `"2 L"`, which means that you have

`x + y = 2" " " "color(orange)((2))`

Now simply use equations `color(orange)((1))` and `color(orange)((2))` to find the values of `x` and of `y`

`x = 2-y`

`0.5 * (2 -y) + 0.1y = 0.40`

`1 - 0.5y + 0.1y = 0.40`

`-0.4y = -0.60 implies y = ((-0.60))/((-0.4)) = 1.5`

This means that `x` is equal to

`x = 2 - 1.5 = 0.5`

So, in order to get `"2 L"` of `"0.2 M"` hydrochloric acid solution, you must mix

`"0.5 L " -> " 0.5 M HCl solution"`

`"1.5 L " -> " 0.1 M HCl solution"`