# Can you help me? From a 0.5 M Na_2CO_3 solution having a volume of 300 ml, prepare a 0.075 M Na_2CO_3 solution with a volume of 150 ml.

Jul 9, 2015

You'd take 22.5 mL of your stock solution and add enough water to get the volume to 150 mL.

#### Explanation:

This time you're dealing with a classic dilution problem.

The idea behind dilution problems is that the number of moles of solute, which in your case is sodium carbonate, must remain constant after the initial solution is diluted.

Dilutions imply that the number of moles of solute remain constant, but that the total volume of the solution increases. This will cause the solution's molarity to decrease.

So, start by figuring out how many moles of sodium carbonate your target solution must contain.

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

${n}_{N {a}_{2} C {O}_{3}} = \text{0.075 M" * 150 * 10^(-3)"L" = "0.01125 moles}$ $N {a}_{2} C {O}_{3}$

This means that the sample you take from the stock solution must contain the exact same number of moles of sodium carbonate.

Knowing the molarity of the stock solution, the volume you'd need is

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

${V}_{\text{stock" = (0.01125cancel("moles"))/(0.5cancel("moles")/"L") = "0.0225 L}}$

This is equivalent to

$0.0225 \cancel{\text{L") * "1000 mL"/(1cancel("L")) = color(green)("22.5 mL}}$

This means that if you take 22.5 mL of your stock solution, and add enough water to make the volume out to 150 mL, you'll get your target solution.

The volume of water needed is

${V}_{\text{water" = 150 - 22.5 = "127.5 mL}}$