Mixing and diluting solutions

A solution forms when, for example, we dissolve ethanol in water. We get a uniform mixture made up of a dissolved substance (the solute, ethanol) and a solvent (water). By a solution we usually mean a liquid solution, but in general we can also have a solution of two solids, for example the alloy bronze.

Concentration of a solution

Concentration tells us how much dissolved substance there is in a solution. There are several ways to give a concentration:

• By mass: how many grams of solute there are in 100 g of solution.
• By volume: how many millilitres of solute there are in 100 ml of solution.
• As a percentage: what percentage of the solution is solute. For example, we can have a 40% solution, which means there are 40 ml of the substance in every 100 ml of solution.

What concentration would a solution of water and ethanol have if we had 500 ml of it, of which 150 ml was ethanol? To get the concentration by volume, we need to know how many ml there are in 100 ml. But we know how many ml there are in half a litre. We can use the unitary method. First we work out how much ethanol there would be in 1 ml of solution, which we do by dividing the amount of ethanol by 500:

$$\frac{150\text{ml}}{500}=0.3\text{ml}$$

Once we know the amount of ethanol in 1 ml, we find the amount in 100 ml by multiplying it by 100:

$$0.3\text{ml}\cdot 100=30\text{ml}$$

So the solution has a concentration of 30 ml of ethanol per 100 ml. It is a 30% solution.

The mixture equation

The mixture equation lets us work out the concentration of the solution we get by mixing two different solutions. Imagine we have two solutions:

1. a 500 ml solution containing 30% ethanol,
2. a 400 ml solution containing 20% ethanol.

What concentration do we get when we mix these solutions? Let V₁ = 500 be the volume of the first solution, V₂ = 400 the volume of the second solution and V = V₁ + V₂ the volume of the resulting solution.

Next we need the volume fraction, which is just the concentration as a percentage written as a fraction. The first solution, with a concentration of 30%, therefore has a volume fraction $c_1=\frac{30}{100}$, and the second solution, with a concentration of 20%, has a volume fraction $c_2=\frac{20}{100}$. (If we worked with grams instead of litres, we would have mass fractions, but the principle would be the same.) Let the resulting solution have a volume fraction c. Now we can write down the mixture equation:

$$c_1\cdot V_1+c_2\cdot V_2=c\cdot V$$

In this equation the only unknown is the final concentration c. We know all the values on the left-hand side from the question, and we also know the final volume V, because it is just the sum of the volumes V₁ and V₂:

$$c_1\cdot V_1+c_2\cdot V_2=c\cdot (V_1+V_2)$$

Let’s substitute the actual values for all the variables:

$$\frac{30}{100}\cdot 500+\frac{20}{100}\cdot 400=c\cdot (500+400)$$

On the left-hand side we now work out the products of the fractions, and on the right-hand side we just add the volumes:

$$150+80=c\cdot 900$$

Now we divide the whole equation by 900, which leaves only the final concentration c on the right-hand side:

$$\frac{150+80}{900}=c$$

We add up the numerator and simply swap the left-hand and right-hand sides of the mixture equation (we typically want the unknown on the left):

$$c=\frac{230}{900}$$

This is our resulting volume fraction. To get it as a decimal, we work out 230 divided by 900:

$$\frac{230}{900}=0.255555\dots$$

We get the percentage by multiplying this ratio by 100, which gives approximately 25.6%. The resulting solution has a concentration of approximately 25.6%, that is, roughly 25.6 ml of ethanol per 100 ml.

What do we add to get the solution we want?

We might also need to answer a slightly different question. Again we have a 500 ml solution containing 30% ethanol, and a second solution containing 55% ethanol. How many ml of the second solution must we mix with the first to get a 40% solution? Again we use the mixture equation

$$c_1\cdot V_1+c_2\cdot V_2=c\cdot (V_1+V_2)$$

The only unknown is V₂; we know everything else. Unlike in the previous case, we know $c=\frac{40}{100}$, because we know what solution we want to end up with. We substitute the known values into the mixture equation:

$$\frac{30}{100}\cdot 500+\frac{55}{100}\cdot V_2=\frac{40}{100}(500+V_2)$$

We multiply the whole equation by 100:

$$30\cdot 500+55\cdot V_2=40\cdot (500+V_2)$$

We work out the product on the left-hand side and expand the brackets on the right-hand side:

$$15000+55\cdot V_2=40\cdot 500+40\cdot V_2$$

We subtract 15,000 from both sides of the equation:

$$55\cdot V_2=40\cdot 500+40\cdot V_2-15000$$

We subtract 40 · V₂ from both sides of the equation:

$$55\cdot V_2-40\cdot V_2=40\cdot 500-15000$$

We add and subtract whatever we can:

$$15\cdot V_2=20000-15000$$

We subtract on the right-hand side:

$$15\cdot V_2=5000$$

We divide both sides of the equation by 15:

$$V_2=\frac{5000}{15}=333.333\dots$$

So we must pour in roughly 333 ml (a third of a litre) of the second, 55% solution to get a solution with a concentration of 40%. We can check that we calculated correctly. The first 500 ml solution contained 30% ethanol, which means it contained 150 ml of ethanol. If we add 333 ml of the second, 55% solution, it contains 183 ml of ethanol. In total we end up with a solution of volume 500 ml + 333 ml, which is 833 ml, containing 150 ml + 183 ml, which is 333 ml of ethanol. When we work out the proportion

$$\frac{333}{833}\approx 0.4$$

we find that we got (after rounding) a 40% solution of ethanol and water.