# 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?

Jul 9, 2015

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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N a O H} = 0.2 \text{moles"/cancel("L") * 200 * 10^(-3)cancel("L") = "0.04 moles}$ $N a O H$

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

$C = \frac{n}{V} \implies V = \frac{n}{C}$

V_"stock" = (0.04 cancel("moles"))/(0.12cancel("moles")/"L") = "0.333 L" = color(red)("333 mL")

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

${n}_{\text{stock" = C * V = 0.12"moles"/cancel("L") * 200 * 10^(-3)cancel("L") = "0.024 moles}}$ $N a O H$

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

${n}_{\text{add" = n_"target" - n_"stock}}$

${n}_{\text{add" = 0.04 - 0.024 = "0.016 moles}}$ $N a O H$

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

0.016cancel("moles") * "40.0 g"/(1cancel("mole")) = "0.64 g"

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.