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Buffer Solutions
Question 1
What is the effect of adding 1 drop of dilute 2 M HCl to 100 mL of pure water?
Solution:
1 drop = 0.05 mL = 0.00005 L
n(HCl) = cV = 2 x 0.00005 = 0.0001mol
This drop is added to 100 mL of water
[H+] = n/V = 0.0001/0.1 = 0.001 mol L-1
pH = -log(0.001) = 3
Thus the pH of pure water changes from 7 to 3 with the addition of 1 drop. The [H+] changed by a factor of 104. Notice that a very small addition of acid had a large effect on the pH.
Question 2
Calculate the pH of the buffer solution containing 0.1 M CH3COOH and 0.1 M CH3COONa.
(Ka for CH3COOH = 1.8 x 10-5)
Solution:
CH3COOH(aq) Á H+(aq) + CH3COO-(aq) (weak acid)
CH3COONa(s) ® Na+(aq) + CH3COO-(aq) (soluble base)
Since CH3COOH is a weak acid [CH3COOH ] is virtually the concentration of the acid in solution. [CH3COOH] = 0.1 M (According to Le Chateliers’ principle, the presence of the CH3COO- ion further restricts the ionisation of the weak acid. thus very little acid ionises.)
Since CH3COONa. is very soluble (a strong electrolyte) all the CH3COONa is ionised as Na+ and CH3COO-. Since CH3COO- is a weak base in the presence of its conjugate acid, the [CH3COO-] is also virtually the concentration of the salt. [CH3COO-] = 0.1 M
Ka = [H+][CH3COO-]
[CH3COOH]
[H+] = Ka [CH3COOH]
[CH3COO-]
= 1.8 x 10-5 x 0.1
0.1
= 1.8 x 10-5
pH = - log [H+] = 4.7447 ~ 4.74
Question 3
Determine the effect on the pH of the buffer solution if one drop of
2 M HCl is added to 100 mL of the buffer solution.
Solution:
Assume all of the H+ ions from the strong acid react with the base in the buffer solution.
n(H+ in 1 drop) = cV = 2 x 0.00005 = 0.0001 mol
(Calculate dilution when 1 drop is added to 100 mL of solution before reaction of the H+ with the base)
[H+] = n/V = 0.0001/0.1 = 0.001 mol L-1
CH3COO-(aq) + H+(aq) ® CH3COOH(aq)
before (0.1) (0.001) (0.1)
after (0.1 - 0.001) (0.1 + 0.001)
(0.099) (0.101)
Ka = [H+][CH3COO-]
[CH3COOH]
[H+] = Ka [CH3COOH]
[CH3COO-]
= 1.8 x 10-3 x 0.101
0.099
= 1.836 x 10-3
pH = - log [H+] = 4.7436 » 4.74
Note the pH has changed very little (4.7447 - 4.7436 = 0.0011) even though one drop in pure water would change the pH from 7 to 3. Thus we can conclude that the solution is an effective buffer.
Buffer Solutions
Read the information in the sheet on buffer solutions provided. Work through the calculations and then use it to answer the questions below.
A buffer solution is one which maintains a fairly constant pH, even when small amounts of acid or alkali are added to it. Such solutions are said to be buffers because they resist significant changes to pH. Buffer solutions are made by mixing a weak acid (e.g. acetic acid) with a salt of the same weak acid (e.g. sodium acetate). That is a weak acid is mixed with the salt of its conjugate base. The presence of the conjugate base reduces the ionisation of the weak acid and similarly, the presence of the acid suppresses the ionisation of the base.
Problems
1. What is a buffer solution? Give some examples of buffer solutions. Why are such solutions important?
2. Identify buffer solutions from the solutions following -
(a) HCl/NaCl (b) HNO2/NaNO2 (c) citric acid/sodium citrate (d) acetic acid/potassium acetate (e) propanoic acid/sodium propanoate (f) sulfuric acid/sodium hydrogensulfate
3. Calculate the [H+] and the pH of a solution that contains 0.01M C6H5COOH (benzoic acid) and 0.03 M C6H5COONa (sodium benzoate). Ka (benzoic acid) = 6.6 x 10-5.
4. Calculate the [H+] and the pH of a solution that contains 0.1M CH3COOH (acetic acid) and 0.2 M CH3COONa (sodium acetate). Use the Ka value in your text.
5. Calculate the [H+] and the pH of a solution that contains 0.2M HCOOH (formic acid) and 0.3 M HCOONa (sodium formate). Ka (formic acid) = 1.8 x 10-4.
6. What is the pH of pure water? Calculate the pH of 100 mL of water to which is added 1 drop of 2 M HCl. (Note: 1 drop = 0.05 mL) Has the pH changed very much from that of pure water?
7. Calculate the new pH of the solution in Question 3 if 1 drop of 2 M HCl is added to 100 mLof the buffer solution. Has the pH changed much compared to the change to pure water. Is this solution an effective buffer solution?
8. Calculate the new pH of the solution in Question 4 if 1 drop of 2 M HCl is added to 100 mLof the buffer solution. Has the pH changed much compared to the change to pure water. Is this solution an effective buffer solution?
9. Calculate the new pH of the solution in Question 5 if 1 drop of 2 M HCl is added to 100 mLof the buffer solution. Has the pH changed much compared to the change to pure water. Is this solution an effective buffer solution?
Indicators - Extra Information
· Indicators are mixtures of a weak acid and its conjugate base where each species has a different colour.
· Each indicator has its own Ka value (being the Ka of its weak acid)
HIn Á H+ + In-
Ka = [H+][In-]
[HIn]
· Indicators change colour over a limited pH range. Each indicator has its own specific range and we must be careful to choose the appropriate indicator for a specific acid-base titration. We choose an indicator for which the pH range of its colour change (end-point) as closely as possible approximates the pH of the equivalence point of the acid/base titration.
e.g. phenolphthalein 8.2 - 10
bromothymol blue 6 - 8
· The average range for most indicators covers approximately two pH units.
· It is therefore also possible to use the Ka of an indicator to determine the pH range at which it changes colour.
· In the middle of its pH range [HIn] = [In-]
\ [In-] = 1
[HIn]
Ka = [H+][In-] = [H+]
[HIn]
· At the middle of the pH range at which the indicator changes colour [H+] = Ka,
or pH = pKa (Note: pKa = - log Ka)
· Since the colour change occurs gradually over an approximate range of two pH units, then the colour change occurs at pH = pKa ± 1
\ if given Ka, find pKa and then pH colour change range = pKa ± 1
Given the following Ka values of several indicators determine the approximate range of pH at which they change colour.
Indicator |
Ka |
pKa |
approx. pH range in which colour change occurs |
type of acid/base titration to which they are suited? |
bromothymol blue |
1 x 10-7 |
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methyl red |
6.3 x 10-6 |
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phenolphthalein |
7.94 x 10--10 |
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alizarin yellow |
1 x 10-11 |
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Buffers - Extra notes
Contain a relatively high concentration of a weak acid and its corresponding weak base in equillibrium;
HA Á H+ + A-
Ka = [H+][In-]
[HA]
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