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:

0dfe0a55af9c8a584ed53a0454d0beea.png

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

1d254f48333b1966906176d57d61030d.png

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 ions

in solution

high percentage forms ions

in solution

Weak

only a small percentage

forms ions in solution

only a small percentage

forms ions in solution

Concentrated

large number of moles

of \(\color{red}{\text{acid}}\) in solution

large number of moles

of \(\color{blue}{\text{base}}\) in solution

Dilute

small number of moles

of \(\color{red}{\text{acid}}\) in solution

small number of moles

of \(\color{blue}{\text{base}}\) 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

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

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