Acids and Bases

Acids

In water, acids give a solution with a pH of less than 7.0.
The common acids are:

When an acid is added to water, the acid releases $H^{+}$ ions into the solution:
$$ HCl_{(g)} + aq \rightarrow H^{+}_{(aq)} + Cl^{-}_{(aq)} $$

The $H^{+}$ ion is the active ingredient which is responsible for acid reactions. An acid is a proton donor.

Bases

Common bases are metal oxides ($MgO$, $CuO$) and metal hydroxides ($NaOH$, $Mg(OH)_{2}$). Ammonia, $NH_{3}$ is also a base. A base is a species that is a proton acceptor.

Alkalis

An alkali is any chemical compound that gives a solution with a pH greater than 7.0 when dissolved in water. Some common alkalis include:

An alkali is a special type of base that dissolves in water forming aqueous $OH^{-}_{(aq)}$ ions:
$$ NaOH_{(s)} + aq \rightarrow Na^{+}_{(aq)} + OH^{-}_{(aq)} $$

In a solution, the hydroxide ions from alkalis neutralise the $H^{+}$ ions from the acid:
$$ H^{+}_{(aq)} + OH^{-}_{(aq)} \rightarrow H_{2}O_{(l)} $$

Ammonia ($NH_{3}$) can dissolve in water to form a weak alkaline solution:
$$ NH_{3(aq)} + H_{2}O_{(l)} \leftrightharpoons NH_{4}^{+} + OH^{-}_{(aq)} $$

Salts

A salt is an ionic compound with the following features:

Acid Salts

Sulfuric acid has two replaceable $H^{+}$ ions and is an example of a diprotic acid. If one $H^{+}$ ion is replaced, an acid salt is formed which itself can behave as an acid as it still contains another $H^{+}$ ion.

Formation of Salts

Salts can be formed in a range of ways:

Ammonium Salts and Fertilisers

Ammonium salts can be used as fertilisers. They are formed when acids react with aqueous ammonia.
$$ NH_{3(aq)} + HNO_{3(aq)} \rightarrow NH_{4}NO_{3(aq)} $$
The ammonium ion $NH^{4+}$ is in place of the metal ion found in common salts.

Water of Crystallisation

Compounds crystallised from water contain water molecules within the resulting crystalline structure. This is called water of crystallisation. When a compound contains water molecules it is hydrated.

For hydrated molecules, the relative number of water molecules is shown after a dot:
$$ CuSO_{4}\cdot5H_{2}O$$

Working out Formula

An experiment was done to determine the formula of hydrated magnesium sulfate:

  • mass of hydrated salt ($MgSO_{4}\cdot xH_{2}O$) = 4.312g
  • mass of anhydrous salt ($MgSO_{4}$) = 2.107g
  • mass of $H_{2}O$ in $MgSO_{4}\cdot xH_{2}O$ = 2.205g
  1. Calculate the amount, in mol of anhydrous $MgSO_{4}$.
    $$ mol = \frac {mass}{M_{r}}
    \hspace3ex
    \frac {2.107}{120.4} = 0.0175\,mol $$
  2. Calculate the moles of $H_{2}O$.
    $$ mol = \frac {mass}{M_{r}}
    \hspace3ex
    \frac {2.205}{18} = 0.1225\,mol $$
  3. Determine the molar ratio.
    $$ MgSO_{4}:
    \frac {0.00175}{0.00175} = 1
    \hspace3ex
    H_{2}O:
    \frac {0.1225}{0.00175} = 7
    \\
    MgSO_{4}\cdot 7H_{2}O
    $$

Moles and Titrations

An acid-base titration is a method of volumetric analysis, where the volume of one solution required to react with another substance is measured. To carry out a titration:

  1. Using a pipette, a measured volume of one solution is added to a conical flask.
  2. The other solution is placed into a burette.
  3. The solution in the burette is added to the solution in the conical flask until the reaction is complete.
  4. This is called the end point of the reaction. At this point, the volume of the solution added from the burette is measured.

To detect the endpoint, an indicator must be used. Below is some possible indicators:

The results from the titration can be used to calculate unknown information about a solution such as concentration, mass or the formula.

Titration: Calculating Concentration

In a titration, $25.0cm^{3}$ of $0.150\,moldm^{-3}$ of potassium hydroxide ($KOH_{(aq)}$) reacted with exactly $34.4cm^{3}$ of sulphuric acid, $H_{2}SO_{4(aq)}$.

$$2KOH_{(aq)} + H_{2}SO_{4(aq)} \rightarrow K_{2}SO_{4(aq)} + H_{2}O_{(l)} $$

  1. Calculate the amount, in mol of $KOH_{(aq)}$ that reacted.
    $$ \eqalign{
    mol &= concentration \times volume \\
    mol &= 0.150 \times 25 \times 10^{-3} \\
    &= 3.75 \times 10^{-3}\,mol} $$
  2. Calculate the moles of $H_{2}SO_{4(aq)}$ used.
    $$ \eqalign{
    2KOH_{(aq)} &+ H_{2}SO_{4(aq)} &\rightarrow K_{2}SO_{4(aq)} &+\,H_{2}O_{(l)} \\
    2\,mol &+ 1\,mol &\rightarrow 1\,mol &+\,1\,mol
    }
    \\
    \
    \
    mol\,of H_{2}SO_{4(aq)} = \frac {3.75 \times 10^{-3}\,mol}{2}
    = 1.875 \times 10^{-3}
    $$
  3. Calculate the concentration of sulphuric acid.
    $$ \frac {1.875 \times 10^{-3}}{34.4 \times 10^{-3}} = 5.45 \times 10^{-2}\,mol\,dm^{-3} $$
© Andrew Deniszczyc, 2018