pH Calulations
pH Calulations Revision
\text{pH} Calculations
\text{pH} is a number that represents the acidity or basicity of an aqueous solution. It is a measure of the concentration of \text{H}^+ ions contained in a solution. \text{pH} is measured on a log scale.
Brønsted Lowry Acids and Bases
- A Brønsted Lowry Acid is a substance that donates protons.
- A Brønsted Lowry Base is a substance that accepts protons.
In the reaction below, the hydrochloric acid acts as an acid as it can donate a proton while the ammonia acts as a base as it can accept protons.
The Brønsted Lowry acid is linked to a conjugate base on the other side of the equation, they are known as a conjugate pair. The conjugate base will likely be a negative ion, capable of accepting a \text{H}+ ion to reform the acid. Similarly, a Brønsted Lowry base is linked to a conjugate acid on the other side of the equation. The conjugate acid will be a positive ion, capable of donating a \text{H}^+ ion to reform the base.
A quick way to identify the conjugate of a substance is that there should be a difference of a proton between them.
In the reaction there is a 1 proton difference between \text{HCl} and \text{Cl}^- so \text{Cl}^- must be the conjugate base. There is also a one proton difference between \text{NH}_3 and \text{NH}_4^+ so \text{NH}_4^+ must be the conjugate acid.
\text{pH} of Strong Acids
Strong acids are acids that completely dissociate in solution.
\text{HX}_{\text{(s)}}\rarr\text{H}^+_{\text{(aq)}}+\text{X}^-_{\text{(aq)}}
As the acid totally dissociates, the \text{pH} of the acid can be calculated directly from the concentration of \text{H}^+ ions. This is done by taking the negative log of the concentration:
\text{pH}=-\text{log}\left[\text{H}^+\right]
When the \text{pH} is known, the concentration of \text{H}^+ ions can be found by following the inverse process:
\left[\text{H}^+\right]=10^{-\text{pH}}
For monoprotic (i.e. acids that only donate one hydrogen ion, such as \text{HCl} or \text{HNO}_3 the concentration of hydrogen ions is the same as the concentration of the acid. For diprotic acids (those that donate two hydrogen ions, such as \text{H}_2\text{SO}_4), the concentration of \text{H}^+ is double that of the acid.
The Ionic Product of Water, Kw
In aqueous solutions, and pure water, \text{H}_2\text{O} will dissociate in the equilibrium below:
\text{H}_2\text{O}_{\text{(l)}}\rightleftharpoons\text{H}^+_{\text{(aq)}}+\text{OH}^-_{\text{(aq)}}
Using the \text{K}_c equation we can create an expression for the equilibrium:
\text{K}_c=\frac{\left[\text{H}^+\right]\left[\text{OH}^-\right]}{\left[\text{H}_2\text{O}\right]}
This can then be rearranged to obtain an expression for the ionic product of water \text{K}_w.
\left[\text{H}^+\right]\left[\text{OH}^-\right]=\text{K}_c\left[\text{H}_2\text{O}\right]=\text{K}_w
\text{K}_c is typically a very small number (almost negligible), while \left[\text{H}_2\text{O}\right] is constant, making \text{K}_w effectively constant. In pure water and neutral solutions \left[\text{H}^+\right] = \left[\text{OH}^-\right] . So the equation can be re-written as
\text{K}_w=\left(\left[\text{H}^+\right]\right)^2
The \text{K}_w value of an aqueous solution at room temperature \left(25\degree \text{C}\right) is 1.0\times10^{-14}\text{ mol}^2\text{ dm}^{-6}.
The ionic product of water can be used to calculate the \text{pH} of a strong base. Strong bases completely dissociate into their ions in solution.
\text{BX}_{\text{(s)}}\rarr\text{B}^-_{\text{(aq)}}+\text{X}^+_{\text{(aq)}}
To calculate the pH of a strong base, we need to use the pH equation as well as the equation for the ionic product of water. This is done by first calculating the value of \left[\text{H}^+\right] by taking the square root of \text{K}_w. This can then be fed into the \text{pH} equation:
\text{pH of a Base}=-\text{log } \sqrt{\text{K}_w}
pH of Weak Acids
Weak acids are ones that only partially dissociate into solution:
\text{HA}_{\text{(s)}}\rightleftharpoons\text{H}^+ + \text{A}^-
This equilibrium can be studied using the acid dissociation constant \text{K}_a. Larger values of \text{K}_a will correspond to compounds that dissociate more completely into solution, and therefore to stronger acids. \text{K}_a is calculated in the same manner as all other equilibrium constants:
\text{K}_a=\frac{\left[\text{H}^+\right]\left[\text{A}^-\right]}{\left[\text{HA}\right]}
This allows the calculation of \left[\text{H}^+\right] and therefore for the \text{pH} of a weak acid. When using \text{K}_a to calculate the pH of weak acids, two key assumptions are made.
- That \left[\text{H}^+\right]=\left[\text{A}^-\right] because monoprotic acids dissociate into a 1:1 ratio of \text{H}^+ ions and conjugate base (A).
- That the concentration of the acid (HA) at equilibrium is the same as the initial concentration of the acid. This is because there is only a small dissociation of the acid and so the actual value of \left[\text{HA}\right] will only be negligibly different from the initial concentration.
These assumptions mean that the Ka expression can be simplified to
\text{K}_a=\frac{\left(\left[\text{H}^+\right]\right)^2}{\left[\text{HA}\right]_{\text{Initial}}}
It is very common that the \text{pK}_a of a substance is given as opposed to its \text{K}_a. \text{pK}_a is the negative logarithm of \text{K}_a, it also represents the strength of an acid. To convert \text{pK}_a to \text{K}_a the following equation can is used:
\text{K}_a=10^{-\text{pK}_a}
Example 1: Calculating the \text{pH} of a strong base
Calculate the \text{pH} of a \textcolor{#00bfa8}{0.15\text{ mol dm}^{-3}} sample of \text{NaOH}. The ionic product of water, \text{K}_w at 25\degree \text{C} is \textcolor{#f21cc2}{1.0\times10^{-14}\text{ mol}^2\text{ dm}^{-6}}.
[3 marks]
Step 1: Calculate the the concentration of \left[\text{H}^+\right].
\begin{aligned}\text{K}_w&=\left[\text{H}^+\right]\left[\text{OH}^-\right]\\\left[\text{H}^+\right]&=\frac{\text{K}_w}{\left[\text{OH}^-\right]}\\ &=\frac{\textcolor{#f21cc2}{10^{-14}}}{\textcolor{#00bfa8}{0.15}}\\ &=\textcolor{#008d65}{6.67\times10^{-14}\text{ mol dm}^{-3}}\end{aligned}
Step 2: Calculate the \text{pH}.
\begin{aligned}\text{pH}&=-\text{log } 6.67\times10^{-14}\\ &=\textcolor{#008d65}{13.2}\end{aligned}
Example 2: Calculating the \text{pH} of a Weak Acid
Propanoic Acid will partially dissociate in water:
\text{CH}_3\text{CH}_2\text{COOH}\rightleftharpoons\text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+
Calculate the \text{pH} of a solution of propanoic acid with a concentration of \textcolor{#00bfa8}{0.04\text{ mol dm}^{-3}}. The \text{K}_a is \textcolor{#f21cc2}{1.34\times10^{-5}\text{ mol dm}^{-3}}.
[3 marks]
Step 1: Calculate the value of \left[\text{H}^+\right].
\begin{aligned}\text{K}_a&=\frac{\left(\left[\text{H}^+\right]\right)^2}{\left[\text{CH}_3\text{CH}_2\text{COOH}\right]}\\\left[\text{H}^+\right]&=\sqrt{\text{K}_a\left[\text{CH}_3\text{CH}_2\text{COOH}\right]}\\ &=\sqrt{\textcolor{#f21cc2}{1.34\times10^{-5}}\times\textcolor{#00bfa8}{0.04}}\\ &=\textcolor{#008d65}{7.32\times10^{-4}\text{ mol dm}^{-3}}\end{aligned}
Step 2: Calculate the \text{pH}.
\begin{aligned}\text{pH}=-\text{log }7.32\times10^{-4}=\textcolor{#008d65}{3.1}\end{aligned}
pH Calulations Example Questions
Question 1: Give the mathematical definitions the following terms:
- \text{K}_w
- \text{pH}
[2 marks]
a. \text{K}_w = \left[\text{OH}^-\right]\left[\text{H}^+\right]
b. \text{pH} = -\text{log } \left[\text{H}^+\right]
Question 2: Calculate the \text{pH} at 25\degree \text{C} of 2.50\text{ mol dm}^{-3} of \text{NaOH}.
[3 marks]
\begin{aligned}\left[\text{H}^+\right]&=\frac{10^{-14}}{2.50}\\ &=4.0\times10^{-15}\text{ mol dm}^{-3}\end{aligned}
\text{pH}=-\text{log }4.0\times10^{-15}=14.4
(One mark per correct line of working.)
Question 3: Explain the term Brønsted Lowry Base and write an equation for the reaction of methylamine \left(\text{CH}_3\text{NH}_2\right) with water to produce an alkaline solution.
[2 marks]
A Brønsted Lowry base is a proton acceptor (/ \text{H}^+ acceptor).
\text{CH}_3\text{NH}_3 + \text{H}_2\text{O}\rightleftharpoons\text{CH}_3\text{NH}_4^+ +\text{OH}^-Question 4: A solution of ethanoic acid has a \text{pH} of 3.2 at 25\degree \text{C}. Given that the \text{K}_a for ethanoic acid is 1.70\times10^{-5}\text{ mol dm}^{-3}.
Calculate the concentration of the acid.
[4 marks]
Step 1: Calculate the value of \left[\text{H}^+\right].
\left[\text{H}^+\right]=10^{-\text{pH}}=6.31\times10^{-4}\text{ mol dm}^{-3}
Step 2: Calculate the value of \left[\text{CH}_3\text{COOH}\right].
\begin{aligned}\text{K}_a&=\frac{\left(\left[\text{H}^+\right]\right)^2}{\text{CH}_3\text{COOH}}\\\left[\text{CH}_3\text{COOH}\right]&=\frac{\left(\left[\text{H}^+\right]\right)^2}{\text{K}_a}\\ &=0.023\text{ mol dm}^{-3}\end{aligned}
(One mark for step 1. Three marks for step 2, one mark per correct line of calculation.)
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