Chemistry calculations can seem very daunting at first glance. Calculating the concentration of ions in a solution, however, is not as hard as it seems.

Ion concentrations can be calculated using the molarity (m) value, which is how many moles (amount) of a substance there are per liter (L) of solution.

Molarity is typically used to refer to the concentration of solute (substance being dissolved) in a solution. Solute can be either an ion or a molecule. Ions are atoms that have an *electrical charge due* to having more or fewer electrons than the *normal atomic number*.

Molecules are made up of two or more atoms, and thus have *different chemical properties* than pure substances such as gases or metals. With both ions and molecules, the *total number* of electrons remain the same after dissolution.

## Assume complete dissociation

Next, we have to assume that all of the potassium ions are removed from the solution. This is not possible in reality, but we must assume it in order to calculate the concentration of potassium ions in the solution.

In order to do this, we have to first know how many moles of potassium are in one liter of the 0.025 M K2CO3 solution. We can do this by converting 0.025 M into molarity and then multiplying that by one liter.

Then, we have to know how many K+ ions are in **one molecule** of K2CO3. By looking this up, we find that there are two K+ **ions per molecule** of K2CO3.

Finally, we *assume complete dissociation* and multiply the number of moles of potassium in one liter by two to get the number of **moles per one molecule** of K2CO3.

## Calculate total volume of solution

Now let’s calculate the total volume of solution. We need to know how much potassium chloride solution we need to make one liter of 0.025 M K2CO3 solution assuming complete dissociation.

We calculated that we need 0.025 mol K2CO3 per liter of solution, so we need 0.025 N K2CO3 per liter of solution. We also calculated that there are 0.025 N K+ ions per N K2CO3, so we have 1 L of solution per 1 L of solid powder.

Therefore, we need to make 1 L of 0.

## Calculate molarity of solution

The next step is to calculate the molarity, or concentration, of your solution. Molarity is defined as how many moles of a substance are in one liter of a solution.

You calculated the mass of your solution, so now you need to find out how many moles of K+ ions are in it. You did this by dividing the mass of K+ ions in your solution by its molecular weight (122 g/mol).

There are approximately twenty-five milliliters (or ounces) in one liter, so you divided your 1.125 L solution by this amount and **got 105 ml** to account for the *lost volume due* to dryness.

Now that you have the mass and volume of your K+ ion solution, you can calculate its concentration! There are 107 moles of K+ ions per 100 mL of solution, so your concentration is 10% ([10%]).

## Divide moles of K+ by initial amount of K+ ions

Now let’s go back to our 0.025 M K2CO3 solution and **assume complete dissociation** of the K+ ions. What is the concentration of K+ ions in this solution?

We know that there are 0.025 moles of K+ ions, so we can write this as 0.025 = N/V, where N is the number of moles and V is the volume. We also know that each ion takes up a certain volume, so we can divide both sides by V to get N=0.025/.025=1mol/0.025L=40molK+.

We can now use the equation NKA+(NKB+)=Nto figure out how many Na+ and KB+ ions are in the solution. There are 40 mol Na+ ions and 40 mol KB+ ions, so there are 80 mol total Na+ and KB+ ions.

## Divide molarity by the number of liquids in the mixture

To figure out the concentration of K+ ions in a 0.025 M K2CO3 solution assuming complete dissociation, you would divide the molarity of K+ by the number of liquids in the mixture.

There are two liquids: water and K2CO3. Since there are no other compounds in the solution, you can assume all of the ions are present as these two components.

The total volume of the solution is one liter, so there are one thousand (1,000) grams of liquid in it. The mass of *1 mol* of a substance is its weight in grams divided by Avogadro’s number, which is 6.**022 × 1023**.

Therefore, 1 kg of *liquid contains 1 mol* of K2CO3.

## Use the equation for calculating the concentration of a single liquid compound

Given that the molar mass of potassium iodide is 302 g/mol and that 0.025 m **potassium iodide solution would contain 0**.025 mol of KI, its density would be approximately 2 g/ml.

Since 0.025 M K2CO3 solution contains 0.025 mol of K+ ions and CO32- ions, its density would be 2 g/ml as well. Since the volume of the solution does not change after adding the solid, we can assume that the concentration of K+ ions in this solution is 2 mmol/L.

Concentration (mM) = Concentration (mM) ÷ Volume (mL) **× 1000 ÷ 1000000 ÷ 100000 ÷ 1000** = 1 × 10−8 M

Therefore, given that complete dissociation occurs, the concentration of K+ ions in a 0.025 M K2CO3 solution assuming complete dissociation is 1×10−8 M.