Buffer Solutions
Buffer Solutions Revision
Buffer Solutions
Buffer solutions are aqueous solutions that can resist significant pH changes when small amounts of acid and alkali are added to them.
Making Acidic and Basic Buffer Solutions
An acidic buffer solution is an aqueous solution made from a weak acid and a salt of the weak acid – for example, propanoic acid and sodium propanoate. The salt of the weak acid (sodium propanoate) is usually made by reacting the weak acid with a strong base. As such Buffer solutions can be made by adding a solid salt, a salt solution or sodium hydroxide to a weak acid.
Basic buffer solutions also follow the same steps, except instead of a weak acid and its salt, a weak base and its salt is used. An example of this would be ammonia and ammonium chloride. The salt (ammonium chloride) is usually formed by reacting the weak base with a strong acid.
In buffer solutions, the concentration of the salt ion is a lot higher than the concentration of the acid/ base.
How Buffers Work
If a small amount of an acid or an alkali is added to a buffer solution, the equilibrium of the buffer solution will shift to oppose the change.
For example, in a propanoic buffer, the following equilibrium is established.
\text{CH}_3\text{CH}_2\text{COOH}_{ \text{ (aq)}} \rightleftharpoons \text{CH}_3\text{CH}_2\text{CO}_2\text{}^{-}_{\text{(aq)}} + \text{H}^{+}\text{ }_{\text{(aq)}}
Upon adding a small amount of acid, the equilibrium would shift to the left to remove the \text{H}^+ ions added. Due to the high salt concentration in the solution, the \text{pH} would remain almost constant.
Alternatively, upon adding a small amount of alkali, \text{OH}^- ions react with \text{H}^+ ions to form water. Since \text{H}^+ ions in the buffer solution are being used up, the equilibrium shifts to the right to increase the amount of \text{H}^+ ions.
Example 1: Calculating the \text{pH} of Buffers Made From Salt Solutions
One way buffer solutions can be made is with a weak acid and a salt solution.
Calculate the \text{pH} of a buffer made from \textcolor{#00bfa8}{55.0\text{ cm}^3} of \textcolor{#f21cc2}{0.150\text{ mol dm}^{-3}} of ethanoic acid and \textcolor{#a233ff}{40.0\text{ cm}^3} of \textcolor{#eb6517}{0.100\text{mol dm}^{-3}} of sodium ethanoate. The \text{K}_a for ethanoic acid is \textcolor{#0040ff}{1.75 \times 10^{-5}}:
[6 marks]
Step 1: Calculate the moles present in the buffer.
\text{Moles of Ethanoic Acid}=\textcolor{#f21cc2}{0.150}\times\frac{\textcolor{#00bfa8}{55.0}}{1000}=\textcolor{#008d65}{8.25 \times 10^{-3}\text{ mol}}
\text{Moles of Sodium Ethanoate}=\textcolor{#327399}{0.100}\times\frac{\textcolor{#a233ff}{40.0}}{1000}=\textcolor{#008d65}{4.00\times10^{-3}\text{ mol}}
Step 2: Use the \text{K}_a equation to work out the concentration of \text{H}^+ ions.
\text{K}_a=\frac{\text{[H}^{+}\text{]}\text{[A}^{-}\text{]}}{\text{[HA]}}
\text{[H}^{+}\text{]}=\frac{\text{K}_{a}\text{[HA]}}{\text{[A}^-\text{]}}
We can use the moles in the \text{K}_a equation because the total volume cancels out.
\text{[H}^{+}\text{]}=\frac{\textcolor{#0040ff}{1.75\times 10^{-5}}\times 8.25\times 10^{-3}}{4\times10^{-3}}=\textcolor{#008d65}{3.61\times10^{-5}\text{ mol dm}^{-3}}
Step 3: Calculate the \text{pH}.
\begin{aligned}\text{pH}&=-\text{log[H}^{+}\text{]}\\ &=-\text{log}(3.61\times 10^{-5})\\&=\textcolor{#008d65}{4.44}\end{aligned}
Example 2: Calculating the \text{pH} of Buffers Made From Solid Salts
One way that buffer solutions can be made is with a weak acid and a solid salt.
Calculate the \text{pH} of a buffer made by adding \textcolor{#00bfa8}{1.70\text{ g}} of sodium ethanoate to \textcolor{#f21cc2}{150\text{ cm}^3} of \textcolor{#a233ff}{0.500\text{mol dm}^{-3}} of ethanoic acid. The \text{K}_a for ethanoic acid is \textcolor{#0040ff}{1.75 \times 10^{-5}}:
[6 marks]
Step 1: Calculate the moles present in the buffer.
\text{Moles of Ethanoic Acid}=\textcolor{#a233ff}{0.500}\times\frac{\textcolor{#f21cc2}{150}}{1000}\textcolor{#008d65}{0.075\text{ mol}}
\text{Moles of Sodium Ethanoate}=\frac{\textcolor{#00bfa8}{1.70}}{84}=\textcolor{#008d65}{0.021\text{ mol}}
Step 2: Use the \text{K}_a equation to work out the concentration of \text{H}^+ ions.
\text{[H}^{+}\text{]}=\frac{\text{K}_{a}\text{[HA]}}{\text{[A}^-\text{]}}
\text{[H}^{+}\text{]}=\frac{\textcolor{#0040ff}{1.75\times 10^{-5}}\times 0.075}{0.021}=\textcolor{#008d65}{6.25\times10^{-5}\text{ mol dm}^{-3}}
Step 3: Calculate the \text{pH}.
\begin{aligned}\text{pH}&=-\text{log[H}^{+}\text{]}\\ &=-\text{log}(6.3309\times 10^{-5})\\ &=\textcolor{#008d65}{4.20}\end{aligned}
Example 3: Calculating the \text{pH} of Buffers Made From Strong Bases.
Buffer solutions can be made with a weak acid and a strong base, such as potassium or sodium hydroxide, which partially neutralises the weak acid.
Calculate the \text{pH} of the resulting buffer solution when \textcolor{#00bfa8}{60.0\text{ cm}^3} of \textcolor{#f21cc2}{0.600\text{mol dm}^{-3}} of propanoic acid reacts with \textcolor{#a233ff}{35.0\text{ cm}^3} of \textcolor{#eb6517}{0.450\text{ mol dm}^{-3}} sodium hydroxide. The \text{K}_a of propanoic acid is \textcolor{#0040ff}{1.35 \times 10^{-5}}:
[8 marks]
Step 1: Determine the species and moles in excess in the buffer.
\begin{aligned}\text{Moles of Propanoic Acid}&=\textcolor{#f21cc2}{0.600} \times \frac{\textcolor{#00bfa8}{60.0}}{1000}\\ &=\textcolor{#008d65}{0.036\text{ mol}}\end{aligned}
\begin{aligned}\text{Moles of Sodium Hydroxide}&=\textcolor{#eb6517}{0.450}\times\frac{\textcolor{#a233ff}{35.0}}{1000}\\ &=\textcolor{#008d65}{0.016\text{ mol}}\end{aligned}
\text{Excess Moles of Propanoic Acid}=0.036-0.016=\textcolor{#008d65}{0.020\text{ mol}}
Step 2: Calculate the concentration of propanoic acid and the concentration of ethanoate ions.
\begin{aligned}\text{[CH}_3\text{CH}_2\text{COOH]}&=\frac{\text{Excess Moles Propanoic Acid}}{\text{Total Volume}}\\\text{}\\ &=\frac{0.020}{0.095}\\\text{}\\ &=\textcolor{#008d65}{0.211\text{ mol dm}^{-3}}\end{aligned}
\begin{aligned}\text{[CH}_3\text{CH}_2\text{COO]}^-&=\frac{\text{Moles of OH}^-\text{ added}}{\text{Total Volume}}\\\text{}\\ &=\frac{0.016}{0.095}\\\text{}\\ &=\textcolor{#008d65}{0.168\text{ mol dm}^{-3}}\end{aligned}
Step 3: Use the \text{K}_a equation to work out the concentration of \text{H}^+ ions.
\begin{aligned}\text{[H}^{+}\text{]}&=\frac{\text{K}_{a}\text{[CH}_3 \text{CH}_2\text{CO}_2\text{]}^-}{\text{[CH}_3\text{CH}_2\text{CO}_2\text{H]}}\\\text{}\\ &=\frac{\textcolor{#0040ff}{1.35\times 10^{-5}}\times 0.211}{0.168}\\\text{}\\ &=\textcolor{#008d65}{1.69\times10^{-5}\text{ mol dm}^{-3}}\end{aligned}
Step 4: Calculate the \text{pH}.
\begin{aligned}\text{pH}&=-\text{log[H}^{+}\text{]}\\ &=-\text{log}(1.69\times 10^{-5})\\ &=\textcolor{#008d65}{4.77}\end{aligned}
Example 4: Calculating \text{pH} Changes in a Buffer
When calculating a buffer solution’s \text{pH} change, the change in moles needs to be considered.
If an acid is added to a buffer solution, the number of moles of the salt in the buffer will decrease while the number of moles of acid in the buffer will increase. Both would change by the number of moles of acid added to the buffer solution.
Similarly, if an alkali is added to a buffer solution, the moles of the salt in the buffer will increase while the moles of acid in the buffer will decrease. Both would change by the amount of moles of alkali added to the buffer solution.
\textcolor{#00bfa8}{0.008\text{ mol}} of sodium hydroxide is added to \textcolor{#f21cc2}{500\text{ cm}^3} of a buffer where the concentration of ethanoic acid is \textcolor{#a233ff}{0.300 \text{mol dm}^{-3}}, and the concentration of sodium ethanoate is \textcolor{#eb6517}{0.200\text{ mol dm}^{-3}}. The \text{K}_a for ethanoic acid is \textcolor{#0040ff}{1.75 \times 10^{-5}}.
Calculate the \text{pH} of the buffer solution after the NaOH has been added:
[8 marks]
Step 1: calculate the amount of moles of ethanoic acid and sodium ethanoate in the initial buffer solution.
\text{Moles of Ethanoic Acid in Initial Buffer}=\textcolor{#a233ff}{0.300} \times \textcolor{#f21cc2}{0.500}=\textcolor{#008d65}{0.150\text{mol}}
\text{Moles of Sodium Ethanoate in Initial Buffer}=\textcolor{#eb6517}{0.200} \times \textcolor{#f21cc2}{0.500}=\textcolor{#008d65}{0.100\text{ mol}}
Step 2: Calculate the final moles present in the buffer.
\text{Final Moles Ethanoic Acid}=0.150-\textcolor{#00bfa8}{0.008}=\textcolor{#008d65}{0.142\text{ mol}}
\text{Final Moles Sodium Ethanoate}=0.100+\textcolor{#00bfa8}{0.008}=\textcolor{#008d65}{0.108\text{ mol}}
Step 3: Use the \text{K}_a equation to work out the concentration of \text{H}^+ ions.
\begin{aligned}\left[\text{H}^+\right]&=\frac{\text{K}_a\left[\text{CH}_3\text{COOH}\right]}{\left[\text{CH}_3\text{COO}^-\right]}\\\text{}\\&=\frac{\textcolor{#0040ff}{1.75\times 10^{-5}}\times 0.142}{0.108}\\ \text{}\\ &=\textcolor{#008d65}{2.30\times 10^{-5}\text{ mol dm}}\end{aligned}
Step 4: Calculate the \text{pH}.
\begin{aligned}\text{pH}&=-\text{log[H}^{+}\text{]}\\ &=-\text{log}(2.30\times 10^{-5})\\ &=\textcolor{#008d65}{4.64}\end{aligned}
Buffer Solutions Example Questions
Question 1: Calculate the pH of the buffer solution formed when 20.0\text{ cm}^{3} of 0.200\text{ mol dm}^{-3} potassium hydroxide is added to 50.0\text{ cm}^{3} of 0.800\text{mol dm}^{-3} ethanoic acid. The Ka of ethanoic acid is 1.73 \times 10^{-5}.
[6 marks]
Step 1: Calculate the initial moles present in the buffer (2 marks).
\text{Moles OH}^-=0.200\times 0.020=4 \times 10^{-3}\text{ mol}
\text{Initial Moles Ethanoic Acid}=0.800\times0.050=0.040\text{ mol}
Step 2: Calculate the final moles present in the buffer (2 marks).
\text{Final Moles Ethanoic Acid}=0.040-(4\times10^{-3})=0.036\text{ mol}
\text{Final Moles Potassium Ethanoate}=\text{Moles OH}^-=4\times10^{-3}
Step 3: Calculate the concentration of \text{[H}^+\text{]} ions (1 mark).
\text{[H}^+\text{]}=\frac{1.73\times10^{-5}\times0.036}{4\times10{-3}}=1.56\times10^{-4}
Step 4: Calculate the \text{pH} (1 mark).
\text{pH}=-\text{log}(1.56\times10^{-5})=3.81
Question 2: Explain why the pH of an acidic buffer solution will remain fairly constant despite the addition of a small amount of sodium hydroxide.
[2 marks]
The additional \text{OH}^- is removed by reaction with \text{H}^+.
The reduction of \text{[H}^+\text{]} is then compensated by the equilibrium of the buffer shifting to the right to produce more.
Question 3: A buffer solution is prepared by dissolving 0.031\text{ mol} of the salt NaY in 45.0\text{ cm}^3 of a 0.431\text{ mol dm}^{-3} solution of the weak acid HY.
Calculate the \text{pH} of this buffer solution. The \text{K}_a of the weak acid is 1.35 \times 10^{-5}.
[3 marks]
Step 1: Calculate the moles of HY:
\text{Moles HY}=0.045\times0.431=\underline{0.019\text{ mol}}
Step 2: Calculate the concentration of \text{[H}^+\text{]} ions.
\text{[H}^+\text{]}=\frac{1.35\times10^{-5}\times 0.019}{0.031}=\underline{8.27\times10^{-6}}
Step 3: Calculate the \text{pH}.
\text{pH}=-\text{log}(8.27\times10^{-6})=\underline{5.08}
(One mark per correct step)
Question 4: A 6.00\times 10^{-4}\text{ mol} sample of sodium hydroxide is added to the buffer solution described in Question 3.
Calculate the pH of this buffer solution after sodium hydroxide is added.
[4 marks]
Step 1: Calculate moles present in the buffer after addition of sodium hydroxide (2 marks).
\text{Moles HY}=0.019-6\times 10^{-4}=\underline{0.0184\text{ mol}}
\text{Moles Y}^-=0.031+6\times 10^{-4}=\underline{0.0314\text{ mol}}
Step 2: Calculate the concentration of \text{[H}^+\text{]} ions (1 mark).
\text{[H}^+\text{]}=\frac{1.35\times10^{-5}\times 0.0184}{0.0316}=\underline{7.96\times10^{-6}}
Step 3: Calculate the \text{pH} (1 mark).
\text{pH}=-\text{log}(7.96\times10^{-6})=\underline{5.10}
Question 5: A buffer solution is prepared containing 2.40\times10^{-2}\text{ mol} of methanoic acid and 1.85\times10^{-2}\text{ mol} of sodium methanoate in 1.00\text{ dm}^3 of solution.
Calculate the \text{pH} of this buffer solution at 25\degree \text{C}. The \text{K}_a of methanoic acid at 25\degree \text{C} is 1.78\times 10^{-4}.
[2 marks]
\text{[H}^+\text{]}=\frac{1.78\times 10^{-4}\times 2.4\times 10^{-2}}{1.85\times 10^{-2}}=\underline{2.31\times 10^{-4}}
\text{pH}=3.64
You May Also Like...
MME Learning Portal
Online exams, practice questions and revision videos for every GCSE level 9-1 topic! No fees, no trial period, just totally free access to the UK’s best GCSE maths revision platform.
