# How would you prepare a 0.2 M #NaOH# solution with a volume of 200 ml, from a 0.12 M #NaOH# solution with a volume of 200 ml?

##### 1 Answer

Here's my take on this problem.

#### Explanation:

Problems that ask you to prepare a solution that has a certain molarity and volume starting from a stock solution can always be approached using the equation for dilution calculations.

When you're *diluting* a solution, you're essentially **increasing** the volume of the solution by keeping the number of moles of solute **constant** and by **increasing** the amount of solvent present.

Since molarity is defined as moles of solute per liters of solution, if you increase the volume *without* changing the number of moles of solute, the molarity will of course *decrease*.

This means that you've successfully diluted the initial solution, i.e. you've made it *less concentrated*.

However, notice that your target solution is actually more **concentrated** than the stock solution - this is inconsistent with the idea that wecan dilute the stock solution to get the target solution.

So, you know that your target solution has a volume of **200 mL** and a molarity of **0.2 M**. This means that it must contain

Now you need to figure out how what volume of the stock solution would contain this many moles of sodium hydroxide.

This is what the problem with the given values is. To get that many moles of sodium hydroxide from that stock solution, you'd need **333 mL**. However, you only have **200 mL** available.

Your stock solution contains

As it turns out, you don't have enough moles of solute to prepare your target solution from your stock solution.

The only way to prepare your target solution and still use that stock solution is to *dissolve* more sodium hydroxide in the stock solution.

You need to add

Use sodium hydroxide's molar mass to determine how many grams would contain this many moles.

If you add **0.64 g** of solid sodium hydroxide to the stock solution you'll get a **0.2-M** solution of (approximately) the same volume.