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Question

A researcher needs a 1.125 M solution of potassium hydroxide. The solution will have a final volume of 250.0 mL. How many grams of potassium hydroxide must be added before the container is filled to its final volume?

Answer 1

If you end with 250 mL of a 1.125 M solution of potassium hydroxide then let's calculate how many moles of potassium hydroxide that is:

So there are 0.28125 moles of potassium hydroxide in that solution. In order to add 0.28125 moles of potassium hydroxide, let's calculate how many grams that is. First I need to know what potassium hydroxide is. Potassium is K and is in the first column on the Periodic Table so has an ionic charge of +1. Hydroxide is the ionic compound OH and has charge of -1. Therefore, they go together as KOH. Next I need to know the molar mass so we calculate that as follows:

K: 39.098 g/mol x 1 = 39.098

O: 15.999 g/mol x 1 = 15.999

H: 1.008 g/mol x 1 = 1.008

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56.105 g/mol

So 0.28125 moles x 56.105 g/mol = 15.7795 grams

Since you are given 4 significant figures in your givens, we round the answer to 4 sig figs: so the mass of potassium hydroxide required is 15.78 g

Answer 2

Hi Kirsten!

This looks like a standard molarity problem. For this, we have to remember that molarity is the measurement of concentration, in the units of moles of solute per liters of solution.

The problem outlines that the desired molarity is 1.125M, which translates to 1.125 moles of potassium hydroxide per 1 liter of solution -- in this case, we have to take into account the fact that we only have 250.0mL of solution, which is equal to 0.2500L.

Molarity = mol KOH / L sol'n

1.125 = mol KOH / 0.2500L sol'n

mol KOH = 0.28125 mol.

Now that we have moles of KOH, we just have to convert from moles to grams using the molar mass of KOH, which is 56.11 g/mol.

Mass KOH = 0.28125 mol * 56.11 g/mol

Mass KOH = 15.78g.

Hope that helps!! If you have any questions, please don't hesitate to ask!