Chemistry » Acid-Base and Redox Reactions » Acids And Bases Continued

# Dilute and Concentrated Solutions

## Dilute and concentrated solutions

A different concept to strong and weak is the concept of concentrated and dilute. Where $$\color{purple}{\text{strong and weak}}$$ refer to the $$\color{purple}{\text{characteristic of a compound}}$$, $$\color{orange}{\text{concentrated and dilute}}$$ refer to the $$\color{orange}{\text{characteristic of a solution}}$$. Thus a strong acid can be prepared as either a concentrated or a dilute solution. A solution of which the exact concentration is known is called a standard solution.

### Definition: Standard solution

A standard solution is one where the exact concentration of solute in a solvent is known.

### Concentrated solutions

A concentrated solution is one where a large amount of a substance (solute) has been added to a solvent. Note that both strong and weak acids and bases can be used in concentrated solutions.

### Definition: Concentrated solution

A concentrated solution is one where there is a high ratio of dissolved substance (e.g. acid or base) to solvent.

### Dilute solutions

A dilute solution is one where a small amount of a substance has been added to a solvent. Note that both strong and weak acids and bases can be used in dilute solutions.

#### Tip:

A concentrated solution has a lot of solute molecules (red circles) in the solvent.

A dilute solution has few solute molecules (red circles) in the solvent.

### Definition: Dilute solution

A dilute solution is one where there is a low ratio of dissolved substance to solvent.

A concentrated solution can be made from a strong or a weak acid or base. A dilute solution can also be made from a strong or a weak acid or base. Whether a solution is concentrated or dilute depends on how much of the acid or base was added to the solvent.

A strong base that is also concentrated would be a base that almost completely dissociates when added to a solution, and you also add a large amount of the base to the solution.

A weak acid that is also dilute would be an acid where only a small percentage of molecules ionise when added to a solution, and you also add only a small amount of the acid to the solution. The table below summarises these concepts.

The electrical conductivity of a solution depends on the concentration of mobile ions in the solution. This means that a concentrated solution of a strong acid or base will have a high electrical conductivity, while a dilute solution of a weak acid or base will have a low electrical conductivity.

#### Fact:

Electric current is the movement of charged particles. Therefore, the more ions (charged particles) there are in a solution, the greater the electric current that can be conducted through the solution by the charged particles. This is the electrical conductivity of a solution.

#### Tip:

Mobile ions are ions that are able to move. These include ions in solution and ions in melted ionic materials. The ions in ionic solids are not mobile.

 $$\color{red}{\textbf{Acid}}$$ $$\color{blue}{\textbf{Base}}$$ Strong high percentage forms ionsin solution high percentage forms ionsin solution Weak only a small percentageforms ions in solution only a small percentageforms ions in solution Concentrated large number of molesof $$\color{red}{\text{acid}}$$ in solution large number of molesof $$\color{blue}{\text{base}}$$ in solution Dilute small number of molesof $$\color{red}{\text{acid}}$$ in solution small number of molesof $$\color{blue}{\text{base}}$$ in solution

Table: A summary of the properties of strong, weak, concentrated, and dilute acids and bases.

## Example:

### Question

Solution 1 contains $$\text{100}$$ $$\text{dm^{3}}$$ of $$\text{HCl}$$ added to $$\text{10}$$ $$\text{dm^{3}}$$ of water. Almost all the $$\text{HCl}$$ molecules ionise in the solution.

Solution 2 contains $$\text{0.01}$$ $$\text{g}$$ of $$\text{Mg}(\text{OH})_{2}$$ added to $$\text{1 000}$$ $$\text{dm^{3}}$$ of water. Only a small percentage of the $$\text{Mg}(\text{OH})_{2}$$ molecules dissociate in the solution.

Say whether these solutions:

1. Contain a strong or weak acid or base.

2. Are concentrated or dilute.

### Step 1: Are the compounds acids or bases?

$$\text{HCl}$$ is hydrochloric acid. It would donate a proton and is an acid. $$\text{Mg}(\text{OH})_{2}$$ is magnesium hydroxide and is a base.

### Step 2: What makes an acid or base strong or weak?

Almost complete ionisation or dissociation means an acid or base is strong. Only a small amount of ionisation or dissociation means an acid or base is weak.

### Step 3: Are the compounds strong or weak acids and bases?

Almost all the $$\text{HCl}$$ molecules ionise in the solution, therefore $$\text{HCl}$$ is a strong acid.

Only a small percentage of the $$\text{Mg}(\text{OH})_{2}$$ molecules dissociate, therefore $$\text{Mg}(\text{OH})_{2}$$ is a weak base.

### Step 4: What makes a solution concentrated or dilute?

A concentrated solution has a high ratio of solute to solvent. A dilute solution has a low ratio of solute to solvent.

### Step 5: Are the solutions concentrated or dilute?

$$\text{100}$$ $$\text{dm^{3}}$$ of $$\text{HCl}$$ is added to $$\text{10}$$ $$\text{dm^{3}}$$ of water. This is a high ratio, therefore the solution of $$\text{HCl}$$ is concentrated.

$$\text{0.01}$$ $$\text{g}$$ of $$\text{Mg}(\text{OH})_{2}$$ is added to $$\text{1 000}$$ $$\text{dm^{3}}$$ of water. This is a low ratio, therefore the solution of $$\text{Mg}(\text{OH})_{2}$$ is dilute.

### Step 6: Combine your information

Solution 1 is a concentrated solution of a strong acid.

Solution 2 is a dilute solution of a weak base.

## Example:

### Question

Solution 1 contains $$\text{0.01}$$ $$\text{dm^{3}}$$ of $$\text{NaOH}$$ added to $$\text{800}$$ $$\text{dm^{3}}$$ of water. Almost all the $$\text{NaOH}$$ molecules dissociate in the solution.

Solution 2 contains $$\text{100}$$ $$\text{g}$$ of $$\text{HF}$$ added to $$\text{10}$$ $$\text{dm^{3}}$$ of water. Only a small percentage of the $$\text{HF}$$ molecules ionise in the solution.

Say whether these solutions:

1. Contain a strong or weak acid or base.

2. Are concentrated or dilute.

### Step 1: Are the compounds acids or bases?

$$\text{NaOH}$$ is sodium hydroxide and is a base. $$\text{HF}$$ is hydrofluoric acid and is an acid.

### Step 2: What makes an acid or base strong or weak?

Almost complete ionisation or dissociation means an acid or base is strong. Only a small amount of ionisation or dissociation means an acid or base is weak.

### Step 3: Are the compounds strong or weak acids and bases?

Almost all the $$\text{NaOH}$$ molecules dissociate in the solution, therefore $$\text{NaOH}$$ is a strong base.

Only a small percentage of the $$\text{HF}$$ molecules ionise, therefore $$\text{HF}$$ is a weak acid.

### Step 4: What makes a solution concentrated or dilute?

A concentrated solution has a high ratio of solute to solvent. A dilute solution has a low ratio of solute to solvent.

### Step 5: Are the solutions concentrated or dilute?

$$\text{0.01}$$ $$\text{g}$$ of $$\text{NaOH}$$ is added to $$\text{800}$$ $$\text{dm^{3}}$$ of water. This is a low ratio, therefore the solution of $$\text{NaOH}$$ is dilute.

$$\text{100}$$ $$\text{dm^{3}}$$ of $$\text{HF}$$ is added to $$\text{10}$$ $$\text{dm^{3}}$$ of water. This is a high ratio, therefore the solution of $$\text{HF}$$ is concentrated.

### Step 6: Combine your information

Solution 1 is a dilute solution of a strong base.

Solution 2 is a concentrated solution of a weak acid.

To calculate the concentration of a solution we use the formula:

$$\text{C (mol.dm}^{-3}{\text{)}}=\dfrac{\text{n (mol)}}{\text{V (dm}^{3}{\text{)}}}$$

## Example:

### Question

$$\text{0.27}$$ $$\text{g}$$ of $$\text{H}_{2}\text{SO}_{4}$$ is added to $$\text{183.7}$$ $$\text{dm^{3}}$$ of water. Calculate the concentration of the solution.

### Step 1: List the information you have and the information you need

V = $$\text{183.7}$$ $$\text{dm^{3}}$$, m = $$\text{0.27}$$ $$\text{g}$$

The volume (V) and the mass (m) are given. The number of moles (n) needs to be calculated. To do that the molar mass (M) needs to be calculated.

### Step 2: Make sure all given units are correct and convert them if necessary

All the units are correct.

### Step 3: What equations will be necessary to calculate the concentration?

$$\text{C (mol.dm}^{-3}{\text{)}}=\dfrac{\text{n (mol)}}{\text{V (dm}^{3}{\text{)}}}$$

$$\text{n (mol)} = \dfrac{\text{m (g)}}{\text{M (g.mol}^{-1}{\text{)}}}$$

### Step 4: Calculate the number of moles of the acid in the solution

M($$\text{H}_{2}\text{SO}_{4}$$) = ($$\text{2}$$ x $$\text{1.01}$$) + $$\text{32.1}$$ + ($$\text{4}$$ x $$\text{16}$$) = $$\text{98.12}$$ $$\text{g.mol^{-1}}$$.

$$\text{n} = \dfrac{\text{m}}{\text{M}} = \dfrac{\text{0.27} {\text{ g}}}{\text{98.12} {\text{ g.mol}}^{-1}} =$$ $$\text{0.0028}$$ $$\text{mol}$$

### Step 5: Calculate the concentration of the solution

$$\text{C} = \dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.0028} {\text{ mol}}}{\text{183.7} {\text{ dm}}^{3}}$$ = $$\text{0.0000152}$$ $$\text{mol.dm^{-3}}$$ = $$\text{1.52} \times \text{10}^{-\text{5}}$$ $$\text{mol.dm^{-3}}$$

## Example:

### Question

$$\text{16.4}$$ $$\text{g}$$ of $$\text{KOH}$$ is added to $$\text{12.9}$$ $$\text{cm^{3}}$$ of water. Calculate the concentration of the solution.

### Step 1: List the information you have and the information you need

V = $$\text{12.9}$$ $$\text{cm^{3}}$$, m = $$\text{16.4}$$ $$\text{g}$$

The volume (V) and the mass (m) are given. The number of moles (n) needs to be calculated. To do that the molar mass (M) needs to be calculated.

### Step 2: Make sure all given units are correct or convert them

The volume needs to be converted to $$\text{dm^{3}}$$.

V = $$\text{12.9}$$ $$\text{cm^{3}}$$ $$\times \dfrac{\text{0.001} {\text{ dm}}^{3}}{\text{1} {\text{ cm}}^{3}}$$ = $$\text{0.0129}$$ $$\text{dm^{3}}$$

### Step : What equations will be necessary to calculate the concentration?

$$\text{C (mol.dm}^{-3}{\text{)}}=\dfrac{\text{n (mol)}}{\text{V (dm}^{3}{\text{)}}}$$

$$\text{n (mol)} = \dfrac{\text{m (g)}}{\text{M (g.mol}^{-1}{\text{)}}}$$

### Step 3: Calculate the number of moles of base in the solution

M($$\text{KOH}$$) = $$\text{39.1}$$ + $$\text{16}$$ + $$\text{1.01}$$ = $$\text{56.11}$$ $$\text{g.mol^{-1}}$$.

$$\text{n}=\dfrac{\text{m}}{\text{M}} = \dfrac{\text{16.4} {\text{ g}}}{\text{56.11} {\text{ g.mol}}^{-1}} =$$ $$\text{0.292}$$ $$\text{mol}$$

### Step 4: Calculate the concentration of the solution

$$\text{C}=\dfrac{\text{n}}{\text{V}} = \dfrac{\text{0.292} {\text{ mol}}}{\text{0.0129} {\text{ dm}}^{3}}$$ = $$\text{22.64}$$ $$\text{mol.dm^{-3}}$$