# Solutions

## Solutions

Earlier, you learnt how to calculate the molar concentration of a solution. The molar concentration of a solution is the number of moles of solute per litre of solvent ($$\text{mol·L^{-1}}$$). This is more commonly given as moles of solute per cubic decimetre of solution ($$\text{mol·dm^{-3}}$$).

To calculate concentration we use $$C = \frac{n}{V}$$, where $$C$$ is the molar concentration, $$n$$ is the number of moles and $$V$$ is the volume of the solution.

Calculating molar concentrations is useful to determine how much solute we need to add to a given volume of solvent in order to make a standard solution.

A standard solution is a solution in which the concentration is known to a high degree of precision. When we work with standard solutions we can take the concentration to be constant.

#### Tip:

When you are busy with these calculations, you will need to remember the following:

$$\text{1}$$ $$\text{dm^{3}}$$=$$\text{1}$$ $$\text{L}$$=$$\text{1 000}$$ $$\text{mL}$$=$$\text{1 000}$$ $$\text{cm^{3}}$$, therefore dividing a volume in $$\text{cm^{3}}$$ by $$\text{1 000}$$ will give you the equivalent volume in $$\text{dm^{3}}$$.

## Example: Concentration Calculations

### Question

How much sodium chloride (in g) will one need to prepare $$\text{500}$$ $$\text{cm^{3}}$$ of a standard solution with a concentration of $$\text{0.01}$$ $$\text{mol·dm^{-3}}$$?

### Step 1: Convert all quantities into the correct units for this equation.

The volume must be converted to $$\text{dm^{3}}$$: \begin{align*} V & = \frac{\text{500}}{\text{1 000}} \\ & = \text{0.5}\text{ dm$^{3}$} \end{align*}

### Step 2: Calculate the number of moles of sodium chloride needed.

\begin{align*} C &= \frac{n}{V} \\ \text{0.01} & = \frac{n}{\text{0.5}} \\ n & = \text{0.005}\text{ mol} \end{align*}

### Step 3: Convert moles of $$\text{NaCl}$$ to mass.

To find the mass of $$\text{NaCl}$$ we need the molar mass of $$\text{NaCl}$$. We can get this from the periodic table (recall from grade 10 how to calculate the molar mass of a compound).

\begin{align*} m & = nM \\ & = (\text{0.005})(\text{58})\\ & = \text{0.29}\text{ g} \end{align*}

The mass of sodium chloride needed is $$\text{0.29}$$ $$\text{g}$$

We will now look at another use of concentration which is for titration calculations.

### Titrations

A titration is a technique for determining the concentration of an unknown solution. Titrations can be done using many different types of reactions. Acid-base reactions and redox reactions are both commonly used for titrations.

In grade 10 you did a simple acid-base titration. Now we will look at how to calculate the concentration of an unknown solution using an acid-base titration.

#### Tip:

When performing a titration we say that the substance of unknown concentration is titrated with the standard solution. A pipette is a measuring device that is used to measure an exact amount of a liquid. If you use a pipette to add liquid to a flask then you would say that the liquid was pipetted into a flask.

We can reduce the number of calculations that we have to do in titration calculations by using the following: \begin{align*} \frac{C_{A}V_{A}}{a} & = \frac{C_{B}V_{B}}{b} \end{align*}

The $$a$$ and $$b$$ are the stoichiometric coefficients of compounds A and B respectively.

## Example: Titration Calculation

### Question

Given the equation:

$$\text{NaOH (aq)} + \text{HCl (aq)} → \text{NaCl (aq)} + \text{H}_{2}\text{O (l)}$$

$$\text{25}$$ $$\text{cm^{3}}$$ of a $$\text{0.2}$$ $$\text{mol·dm^{-3}}$$ hydrochloric acid solution was pipetted into a conical flask and titrated with sodium hydroxide. It was found that $$\text{15}$$ $$\text{cm^{3}}$$ of the sodium hydroxide was needed to neutralise the acid. Calculate the concentration of the sodium hydroxide.

### Step 1: Write down all the information you know about the reaction, and make sure that the equation is balanced.

$$\text{NaOH}$$: $$V =\text{15}\text{ cm^{3}}$$

$$\text{HCl}$$: $$V=\text{25}\text{ cm^{3}}$$; $$C= \text{0.2}\text{ mol·dm^{-3}}$$

### Step 2: Convert the volume to $$\text{dm^{3}}$$

\begin{align*} V_{\text{NaOH}} & = \frac{\text{15}}{\text{1 000}} \\ & = \text{0.015}\text{ dm$^{3}$} \end{align*}\begin{align*} V_{\text{HCl}} & = \frac{\text{25}}{\text{1 000}} \\ & = \text{0.025}\text{ dm$^{3}$} \end{align*}

### Step 3: Calculate the concentration of the sodium hydroxide

\begin{align*} \frac{C_{A}V_{A}}{a} & = \frac{C_{B}V_{B}}{b} \\ \frac{(\text{0.2})(\text{0.025})}{\text{1}} & = \frac{(C_{\text{NaOH}})(\text{0.015})}{\text{1}} \\ \text{0.005} & = (\text{0.015})C_{\text{NaOH}}\\ C_{\text{NaOH}} & = \text{0.33}\text{ mol·dm$^{-3}$} \end{align*}

The concentration of the $$\text{NaOH}$$ solution is $$\text{0.33}$$ $$\text{mol·dm^{-3}}$$

## Example: Titration Calculation

### Question

$$\text{4.9}$$ $$\text{g}$$ of sulfuric acid is dissolved in water and the final solution has a volume of $$\text{220}$$ $$\text{cm^{3}}$$. Using an acid-base titration, it was found that $$\text{20}$$ $$\text{cm^{3}}$$ of this solution was able to completely neutralise $$\text{10}$$ $$\text{cm^{3}}$$ of a sodium hydroxide solution. Calculate the concentration of the sodium hydroxide in $$\text{mol·dm^{-3}}$$.

### Step 1: Write a balanced equation for the titration reaction.

$$\text{H}_{2}\text{SO}_{4}\text{(aq)} + 2\text{NaOH (aq)} → \text{Na}_{2}\text{SO}_{4}\text{(aq)} + 2\text{H}_{2}\text{O (l)}$$

### Step 2: Calculate the concentration of the sulfuric acid solution.

First convert the volume into $$\text{dm^{3}}$$:

\begin{align*} V & = \frac{\text{220}}{\text{1 000}} = \text{0.22}\text{ dm$^{3}$} \end{align*}

Then calculate the number of moles of sulfuric acid:

\begin{align*} n & = \frac{m}{M} \\ & = \frac{\text{4.9}}{\text{98}}\\ & = \text{0.05}\text{ mol} \end{align*}

Now we can calculate the concentration of the sulfuric acid:

\begin{align*} C & = \frac{n}{V} \\ & = \frac{\text{0.05}}{\text{0.22}}\\ & = \text{0.227}\text{ mol·dm$^{-3}$} \end{align*}

### Step 3: Calculate the concentration of the sodium hydroxide solution.

Remember that only $$\text{20}$$ $$\text{cm^{3}}$$ or $$\text{0.02}$$ $$\text{dm^{3}}$$ of the sulfuric acid solution is used.

\begin{align*} \frac{C_{1}V_{1}}{n_{1}} & = \frac{C_{2}V_{2}}{n_{2}} \\ \frac{(\text{0.227})(\text{0.02})}{\text{1}} & = \frac{(C_{\text{NaOH}})(\text{0.01})}{\text{2}} \\ \text{0.00454} & = (\text{0.005})C_{\text{NaOH}}\\ C_{\text{NaOH}} & = \text{0.909}\text{ mol·dm$^{-3}$} \end{align*}