# Question #f9869

##### 1 Answer

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

The **number of moles** of hydrochloric acid coming from solution

#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

#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** **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

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

Now simply use equations

#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 = 2 - 1.5 = 0.5#

So, in order to get

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

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