# Diluting and mixing solutions

A solution is formed when, for example, we dissolve ethanol in water. This produces a homogeneous mixture that consists of the solute (ethanol) and the solvent (water). Usually by solution we mean a solution in the liquid state, but in general we can also have a solution of two solids, for example an alloy of bronze.

## The concentration of the solution

Concentration tells us how much solute is in a solution. We have several ways of giving the concentration:

**Mass:**how many grams of solute are in 100 grams of solution.**Volumetric:**how many millilitres of solute are in 100 ml of solution.**Proportional:**what percentage of the solute is in the solution, for example we may have a 40% solution, which means there is 40 ml of solute per 100 ml of solution.

What would be the concentration of a solution of water and ethanol if we had 500 ml of this solution, 150 ml of which would be ethanol? To get the concentration by volume, we need to know how many ml are contained in 100 ml. But we know how many ml are contained in half a litre. We can use the trinomial to help us. First we calculate how much ethanol would be in 1 ml of the solution, which we do by dividing the amount of ethanol by 500:

$$\large\frac{150\ \mbox{ml}}{500}=0{,}3\ \mbox{ml}$$

When we know the amount of ethanol in 1 ml, we find the amount in 100 ml by multiplying that amount by 100:

$$\large0{,}3\ \mbox{ml}\cdot100=30\ \mbox{ml}$$

The solution thus has a concentration of 30 ml of ethanol per 100 ml. This is a 30% solution.

## The mixing equation

Using the mixing equation, we can calculate the concentration of the solution obtained by mixing two different solutions. Let us imagine that we have two solutions:

- 500ml solution that contains 30% ethanol,
- a 400ml solution containing 20% ethanol.

What is the concentration of the solution we get when we mix these solutions? Let's label V_{1} = 500 the volume of the first solution, V_{2} = 400 the volume of the second solution, and V = V_{1} + V_{2} the volume of the resulting solution.

Next, we will need the so-called volume fraction, which is just a percentage of the concentration converted to fraction form. So the first solution of 30% concentration has a volume fraction of $c_1 = \frac{30}{100}$ and the second solution of 20% concentration has a volume fraction of $c_2 = \frac{20}{100}$. (If we were working with grams instead of liters, we would have mass fractions, the principle would be the same.) Let the resulting solution have the volume fraction c. Now we can write down the mixing equation:

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

In this equation, the only unknown for us is the resulting concentration c. We know all the values to the left of the equation from the assignment, and we know the resulting volume V as well, it's just the sum of the volumes V_{1} and V_{2}:

$$\large c_1\cdot V_1+c_2\cdot V_2=c\cdot (V_1 + V2)$$

Let's add the specific values for all the variables:

$$\frac{30}{100}\cdot500+\frac{20}{100}\cdot400=c\cdot(500+400)$$

On the left side, we now calculate the sums of the fractions, and on the right side, we just add the volumes:

$$\large150+80=c\cdot900$$

Now divide the whole equation by 900, which leaves us with only the final concentration on the right side c:

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

We add the numerator and just swap the left and right sides of the mixing equation (typically we want the unknown on the left):

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

This is our final volume fraction. If we want to get a ratio, we calculate 230 divided by 900:

$$\large\frac{230}{900}=0{,}255555…$$

We get the percentage expression by multiplying this ratio by 100 to get approximately 25.5%. The resulting solution will have a concentration of approximately 25.5% or have approximately 25.5 ml of ethanol per 100 ml.

## What should we dilute to get the desired solution?

But we may also need to find out a slightly different question. Again, let's have a 500ml solution that contains 30% ethanol and a second solution that contains 55% ethanol. How many ml of the second solution do we need to mix with the first to get a 40% solution? Again, we use the mixing equation

$$\large c_1\cdot V_1+c_2\cdot V_2=c\cdot (V_1 + V2)$$

The only unknown for us is V_{2}, we know everything else. Unlike the previous case, we know $c = \frac{40}{100}$, because we know what solution we want to mix. We plug the known values into the mixing equation:

$$\large \frac{30}{100}\cdot500+\frac{55}{100}\cdot V_2=\frac{40}{100}(500+V_2)$$

Multiply the whole equation by 100:

$$\large 30\cdot500+55\cdot V_2=40\cdot(500+V_2)$$

Calculate the product on the left side and multiply the parenthesis on the right side:

$$\large 15\ 000+55\cdot V_2=40\cdot500+40\cdot V_2$$

Subtract 15 000 from both sides of the equation:

$$\large 55\cdot V_2=40\cdot500+40\cdot V_2-15\ 000$$

Subtract 40 · V_{2} from both sides of the equation:

$$\large 55\cdot V_2-40\cdot V_2=40\cdot500-15\ 000$$

Add/subtract what we can:

$$\large 15\cdot V_2=20\ 000-15\ 000$$

Subtract the right side:

$$\large 15\cdot V_2=5\ 000$$

Divide both sides of the equation by 15:

$$\large V_2=\frac{5\ 000}{15}=333{,}333…$$

So, we need to add about 333 ml (one-third of a liter) of the second 55% solution to get a 40% solution. We can check that we have calculated correctly. The first 500ml solution contained 30% ethanol, which means it contained 150ml of ethanol. If we add 333 ml of the second 55% solution, it contains 183 ml of ethanol. As a result, we got a solution of 500 ml + 333 ml, which is 833 ml, which contains 150 ml + 183 ml which is 333 ml of ethanol. If we calculate the proportion

$$\frac{333}{833}\approx 0{,}4$$

We got (after rounding) a 40% solution of ethanol and water.