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Strength of acids and bases quiz
\[\require{mhchem}\]
Worked Exercises
The weak acid \(\ce{HA}\) in \(0.0200M\) solution is \(10\%\) ionised. Calculate the acid ionisation constant for the acid.
Answer
We are given the percentage ionisation of the weak acid and asked to calculate the acid ionisation constant for the acid. First, we need to calculate the concentrations of \(\ce{H}_{3}\ce{O}^{+}\), \(\ce{A}^{-}\) and \(\ce{HA}\) using the percentage ionisation. The ionisation reaction can be written as follows: \[ \ce{HA}\left(aq\right)+\ce{H}_{2}\ce{O}\left(l\right)\rightleftharpoons\ce{H}_{3}\ce{O}^{+}\left(aq\right)+\ce{A}^{-}\left(aq\right) \]
The concentration of \(\ce{H}_{3}\ce{O}^{+}\) will be the \(10\%\) of the original concentration of \(\ce{HA}\). \[ \frac{0.0200M}{100}\times10=0.00200M \]
As the stoichiometric ratio between \(\ce{H}_{3}\ce{O}^{+}\) and \(\ce{A}^{-}\) is one to one \(1:1\), the concentration of \(\ce{A}^{-}\)will also be \(0.00200M\).
The remaining concentration of \(\ce{HA}\) is: \[ 0.0200M-0.00200M=0.0180M \]
Next, substitute the concentration values into the acid ionisation constant expression: \[\begin{align*} K_{a} & =\frac{\left[\ce{H}_{3}\ce{O}^{+}\right]\left[\ce{A}^{-}\right]}{\left[\ce{HA}\right]}\\ K_{a} & =\frac{0.00200M\times0.00200M}{0.0180M}\\ K_{a} & =2.2\times10^{-4} \end{align*}\]
The strength of a base can also be predicted by the ionisation constant, which is known as base ionisation/dissociation constant \(K_{b}\). The general expression for \(K_{b}\) is: \[\begin{align*} \ce{B}\left(aq\right)+\ce{H}_{2}\ce{O}\left(l\right) & \rightleftharpoons\ce{BH}^{+}\left(aq\right)+\ce{OH}^{-}\left(aq\right)\\ K_{b} & =\frac{\left[\ce{OH}^{-}\right]\left[\ce{BH}^{+}\right]}{\left[\ce{B}\right]} \end{align*}\]
Exercises
-
The \(\ce{HBr}\) in a \(0.1M\) \(\ce{HBr}\) solution is \(100\%\) dissociated. What are the molar concentrations of \(\ce{HBr}\), \(\ce{H}_{3}\ce{O}^{+}\) and \(\ce{Cl}^{-}\) in the solution?
\(\ce{H}_{3}\ce{O}^{+}\) \(=0.1M\), \(\ce{Br}^{-}\) \(=0.1M\), \(\ce{HBr}\) \(=0\)
-
How many acidic hydrogens and non-acidic hydrogens are present in the following molecules?
- \(\ce{HNO}_{3}\)
- \(\ce{H}_{2}\ce{C}_{4}\ce{H}_{4}\ce{O}_{4}\) (succinic acid)
- \(\ce{NH}_{3}\)
- \(\ce{CH}_{4}\)
| Acidic | Non-acidic | |
|---|---|---|
| a | \(1\) | \(0\) |
| b | \(2\) | \(4\) |
| c | \(3\) | \(0\) |
| d | \(0\) | \(4\) |
- Using the acid ionisation constants given below, indicate which acid is the stronger in each of the following acid pairs.
| Formula | \(K_{a}\) |
|---|---|
| \(\ce{H}_{3}\ce{PO}_{4}\) | \(7.5\times10^{-3}\) |
| \(\ce{HF}\) | \(6.8\times10^{-4}\) |
| \(\ce{HNO}_{2}\) | \(4.5\times10^{-4}\) |
| \(\ce{HCN}\) | \(4.9\times10^{-10}\) |
| \(\ce{H}_{2}\ce{CO}_{3}\) | \(4.3\times10^{-7}\) |
| \(\ce{HCO}_{3}^{-}\) | \(5.6\times10^{-11}\) |
- \(\ce{H}_{3}\ce{PO}_{4}\) and \(\ce{HNO}_{2}\)
- \(\ce{HCN}\) and \(\ce{HF}\)
- \(\ce{H}_{2}\ce{CO}_{3}\) and \(\ce{HCO}_{3}^{-}\)
- \(\ce{HNO}_{2}\) and \(\ce{HCN}\)
- \(\ce{H}_{3}\ce{PO}_{4}\) is stronger
- \(\ce{HF}\) is stronger
- \(\ce{H}_{2}\ce{CO}_{3}\) is stronger
- \(\ce{HNO}_{2}\) is stronger
4. A \(0.0300M\) of a base is \(7.5\%\) ionised. Calculate the base ionisation constant \(K_{b}\).
The general equation for a base ionisation in water is: \[ \ce{B}\left(aq\right)+\ce{H}_{2}\ce{O}\left(l\right)\rightleftharpoons\ce{BH}^{+}\left(aq\right)+\ce{OH}^{-}\left(aq\right) \] The concentration of the \(\ce{OH}^{-}\) will be the \(7.5\%\) of the original concentration of \(\ce{B}\). \[ \frac{0.0300M}{100}\times7.5=0.00225M \]
As the stoichiometric ratio between \(\ce{OH}^{-}\) and \(\ce{BH}^{+}\)is \(1:1\) the concentration of \(\ce{BH}^{+}\) will also be \(0.00225M\).
The remaining concentration of \(\ce{B}\) is: \[ 0.0300M-0.00225M=0.0278M \]
Next, substitute the concentration values into the acid ionisation constant expression: \[\begin{align*} K_{a} & =\frac{\left[\ce{OH}^{-}\right]\left[\ce{BH}^{+}\right]}{\left[\ce{B}\right]}\\ K_{a} & =\frac{0.00225M\times0.00225M}{0.0278M}\\ K_{a} & =1.82\times10^{-4} \end{align*}\]
5. A \(0.0500M\) of an acid is \(15\%\) ionised. Calculate the acid ionisation constant \(K_{a}\).
The ionisation reaction for an acid can be written as follows: \[ \ce{HA}\left(aq\right)+\ce{H}_{2}\ce{O}\left(l\right)\rightleftharpoons\ce{H}_{3}\ce{O}^{+}\left(aq\right)+\ce{A}^{-}\left(aq\right) \]
The concentration of the \(\ce{H}_{3}\ce{O}^{+}\)will be the \(15\%\) of the original concentration of \(\ce{HA}\): \[ \frac{0.0500M}{100}\times15=0.00750M \]
As the stoichiometric ratio between \(\ce{H}_{3}\ce{O}^{+}\) and \(\ce{A}^{-}\)is \(1:1\) the concentration of \(\ce{A}^{-}\)will also be \(0.00750M\).
The remaining concentration of \(\ce{HA}\) is: \[ 0.0500M-0.00750M=0.0425M \]
Next, substitute the concentration values into the acid ionisation constant expression: \[\begin{align*} K_{a} & =\frac{\left[\ce{H}_{3}\ce{O}^{+}\right]\left[\ce{A}^{-}\right]}{\left[\ce{HA}\right]}\\ K_{a} & =\frac{0.00750M\times0.00750M}{0.0425M}\\ K_{a} & =1.32\times10^{-3} \end{align*}\]