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Titration curves
A titration is a procedure for carrying out a chemical reaction between two solutions by the controlled addition from a buret of one solution (the titrant) to the other, allowing measurements to be made throughout the reaction. For a reaction between an acid and a base, a titration is useful for measuring the pH at various points throughout the reaction.
A titration curve is a graph of the pH as a function of the amount of titrant (acid or base) added.
Strong Acid-Strong Base Titrations
Here is an example of a titration curve, produced when a strong base is added to a strong acid. This curve shows how pH varies as 0.100 M NaOH is added to 50.0 mL of 0.100 M HCl.
The equivalence point of the titration is the point at which exactly enough titrant has been added to react with all of the substance being titrated with no titrant left over. In other words, at the equivalence point, the number of moles of titrant added so far corresponds exactly to the number of moles of substance being titrated according to the reaction stoichiometry. (In an acid-base titration, there is a 1:1 acid:base stoichiometry, so the equivalence point is the point where the moles of titrant added equals the moles of substance initially in the solution being titrated.)
Notice that the pH increases slowly at first, then rapidly as it nears the equivalence point. Why?
Learning Goal 25 Calculate the pH at any point, including the equivalence point, in an acid-base titration.
At the equivalence point, the pH = 7.00 for strong acid-strong base titrations. However, in other types of titrations, this is not the case.
EXAMPLE:
What is the pH when 49.00 mL of 0.100 M NaOH solution have been added to 50.00 mL of 0.100 M HCl solution?
Because it is a strong acid-base reaction, the reaction simplifies to:
H+^ (aq) + OH-^ (aq) H 2 O (l)
The original number of moles of H+^ in the solution is:
50.00 x 10-3^ L x 0.100 M HCl = 5.00 x 10-3^ moles
The number of moles of OH-^ added is :
49.00 x 10-3^ L x 0.100 M OH-^ = 4.90 x 10-3^ moles
Thus there remains:
(5.00 x 10-3) - (4.90 x 10-3) = 1.0 x 10-4^ moles H+^ (aq)
The total volume of solution is 0.04900 L + 0.05000 L = 0.09900 L
[H+] = {1.0 x 10-4^ moles / 0.09900 L } = 1.0 x 10-3^ M pH = 3.
Titrations Involving a Weak Acid or Weak Base
Titration curve of a weak acid being titrated by a strong base:
Similarly, there are 3.00 x 10-3^ moles of OH-^ due to the NaOH solution.
The reaction goes to completion:
OH-^ (aq) + HC 2 H 3 O 2 (aq) C 2 H 3 O 2 -^ (aq) + H 2 O (l) OH-^ HC 2 H 3 O 2 C 2 H 3 O 2 - INITIAL 3.00 x 10-3^ mol 5.00 x 10-3^ mol 0 CHANGE -3.00 x 10-3^ mol -3.00 x 10-3^ mol +3.00 x 10-3^ mol FINAL 0 2.00 x 10-3^ mol 3.00 x 10-3^ mol
The total volume is 80.0 mL.
We now calculate the resulting molarities :
[HC 2 H 3 O 2 ] = { 2.00 x 10-3^ mol HC 2 H 3 O 2 / 0.0800 L } = 0.0250 M [C 2 H 3 O 2 - ] = { 3.00 x 10-3^ mol C 2 H 3 O 2 -^ } / 0.0800 L } = 0.0375 M
STEP 2: Equilibrium calculation, using simplification:
Ka = { [H+][C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] } = 1.8 x 10- [H+] = { KA [HC 2 H 3 O 2 ] / [C 2 H 3 O 2 - ] } = { (1.8 x 10-5)(0.0250) / (0.0375) } = 1.2 x 10-5^ M
pH = -log(1.2 x 10-5) = 4.
Titration curve of a weak base being titrated by a strong acid:
Here, 0.100 M HCl is being added to 50.0 mL of 0.100 M ammonia solution.
As in the weak acid-strong base titration, there are three major differences between this curve (in blue) and a strong base-strong acid one (in black): (Note that the strong base-strong acid titration curve is identical to the strong acid-strong base titration, but flipped vertically.)
- The weak-acid solution has a lower initial pH.
- The pH drops more rapidly at the start, but less rapidly near the equivalence point.
- The pH at the equivalence point does not equal 7.00.
POINT OF EMPHASIS : The equivalence point for a weak base-strong acid titration has a pH < 7.00.
Titrations of Polyprotic Acids
An example of a polyprotic acid is H 2 CO 3 which neutralizes in two steps:
H 2 CO 3 (aq) + OH-^ (aq) H 2 O (l) + HCO 3 -^ (aq)
HCO 3 -^ (aq) + OH-^ (aq) H 2 O (l) + CO 3 2-^ (aq)
The titration curve for these reactions will look like this, with two equivalence points.
Uses of Titrations
Learning Goal 26 Use titration data or a titration curve to calculate reaction quantities such as the concentration
Certain chemicals have the special property of appearing one colour when in a solution of one pH and a different colour when in a solution of a different pH. Such chemicals are known as acid-base indicators , or simply as indicators because, when a few drops of indicator are added to a solution, the colour of the solution serves as an indication of its pH.
Most acid-base indicators are simply weak acids with a protonated (acidic) form that is one colour and a deprotonated (basic) form that is a different colour. Like all weak acids, they dissocate according to their Ka value. For a generic indicator HIn,
HIn(aq) H+(aq) + In-(aq) so Ka= { [H+][X-] / [HX] }
Converting this equation into the form of the Henderson-Hasselbalch equation, we get
pH = pKa + log([In-]/[HIn])
This shows us that the pH of the solution and the pKa of the indicator together determine the ratio of In-^ to HIn. Since the two species are different colours, this ratio in turn determines the overall colour of the solution.
EXAMPLE:
Consider the indicator phenol red, which has a yellow HIn form, a red In-^ form, and a Ka of 5.0x10-8. Now imagine that a few drops of this indicator are added to a solution of pH 2.3. What colour would the solution be?
To figure this out, we need the above equation:
pH = pKa + log([In-]/[HIn]) pH = -log(Ka) + log([In-]/[HIn]) 2.3 = 7.3 + log([In-]/[HIn]) -5.0 = log([In-]/[HIn]) 1.0x10-5^ = [In-]/[HIn]
Thus, the ratio of In-^ to HIn is 1 to 100 000. Since In-^ is red and HIn is yellow, there are 100 000 yellow molecules in solution for every red molecule. For coloured species, this is an enormous ratio, and the red molecules will be undetectable. Thus, the solution will appear yellow.
EXAMPLE:
Imagine we have a solution of acetic acid of unknown concentration. How could we measure the concentration of the solution using only indicators?
STEP 1: Use the indicators to estimate the pH of the solution
To do this, we remove a small amount of the solution and add a drop of indicator. Based on the resulting colour of the solution, we will know whether the pH is below, within or above the pH range in which the indicator changes colour. We could just do this with several random indicators and hope that they will provide enough information for us. However, we will use a more systematic approach.
Since we do not know the pH of our solution, an indicator that changes colour in the middle of the pH range is a good one to start with since it will split the pH range into roughly equal parts, each significantly smaller than the whole range. Let's say we add a drop of phenol red and the solution turns yellow. Since phenol red is yellow in solutions below pH 6.6, we know that the pH of our acetic acid solution is below 6.6.
Now we can narrow down the pH range with another indicator. A good choice would be one that will further narrow down the pH range that we know contains the solution's pH. For example, let's say we add a drop of methyl red to a new sample of the solution and it turns red. Now we know that the pH is below 4.8, which gives us more information than we had before. In contrast, if we had used alizarin yellow, it would have turned yellow, telling us that the pH is below 10.1. Since the phenol red told us that the pH is below 6.6, the alizarin yellow test gives us no new information.
Using the results of the phenol red and methyl red tests, our third indicator needs to have a colour-change range even lower than 4.8. Suppose we use thymol blue, and it turns yellow. This tells us that the pH of the solution is above 2.8 AND that it is below 8.0. (Since thymol blue changes colour twice, the colour tells us where the pH of the solution lies relative to both of the pKa's of thymol blue.) So, we now know that the pH is between 2.8 and 4.8.
Suppose we now use methyl orange as a final indicator and it turns yellowish orange. This is an intermediate colour for methyl orange, meaning that the pH is within its colour-change range of 3.2 to 4.4. Since it is closer to the yellow end of the range than the red end, we can assume that the pH is in the upper half of the range, namely between 3.8 and 4.4.
Unfortunately, none of our indicators has a range with one endpoint within this range. Thus, additional tests will not give us any more conclusive information, so we have locallized the pH as much as possible.
STEP 2: Convert the pH range into a range of H+^ concentrations:
Using [H+] = 10-pH, we find that a pH range of 3.8 to 4.4 corresponds to an H+^ concentration range of 4.0x10-5M to 1.6x10-4M.
STEP 3: Use an I.C.E. table and the Ka of acetic acid to convert the H+^ concentration range to a range of acetic acid concentrations. 4.0x10-5M:
HC 2 H 3 O 2 H+^ C 2 H 3 O 2 -
INITIAL x 0 0 CHANGE -4.0x10-5M +4.0x10-5M +4.0x10-5M EQUILIBRIUM (x-4.0x10-5)M 4.0x10-5M 4.0x10-5M
Ka = { [H+][C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] }
1.8x10-5^ = { (4.0x10-5)(4.0x10-5) / (x - 4.0x10-5) }
x = 1.3x10-4M = [HC 2 H 3 O 2 ]
For a strong acid-strong base titration, the equivalence point occurs at pH 7.
When an acid is being titrated with a base, the first observable colour change will occur when the In-^ to HIn ratio is approximately 1 to 10.
pH = pKa + log([In-]/[HIn]) 7 = pKa + log(1/10) 7 = pKa - 1 8 = pKa
When a base is being titrated with an acid, the first observable colour change will occur when the In-^ to HIn ratio is approximately 10 to 1.
pH = pKa + log([In-]/[HIn]) 7 = pKa + log(10) 7 = pKa + 1 6 = pKa
So, we see that we actually want different indicators for the two titrations.
In practice, the first colour change does not necessarily have to happen right at the equivalence point. Since we only care about being able to use the colour change to detect when the equivalence point has been reached, the crucial requirement for choosing a good indicator is that it begins to change colour after the volume of titrant required to reach the quivalence point has been added.
For example, a strong acid-strong base has a very sharp equivalence point, meaning that a very large change in pH occurs due to the addition of a very small amount of titrant, often a single drop. Therefore, any pH in this range will be reached after the addition of the same number of drops. This means that any indicator that starts to change colour in this range will signal equally well that the equivalence point has been reached. It also means that the pH range over which the indicator changes colour will likely be passed in a single drop of titrant, resulting in a very sharp endpoint.
A titration involving a weak acid or base, however, has a less steep equivalent point with a less dramatic pH change. Thus, there is a smaller range of pH's that will be passed at the same time as the pH as the equivalence point, meaning that the chosen indicator must begin to change colour closer to the equivalence point than in a strong acid-strong base titration, and that the endpoint will be wider.
Here are the titrations of both a strong acid and a weak acid with a strong base, using methyl red and thymol blue as possible indicators.
In the strong acid titration, both indicators begin to change colour at the equivalence point ( mL of base) so both work equally well. In the weak acid titration, thymol blue changes colour at the equivalence point, but methyl red begins to change colour after only 15mL of base are added, which is far from the equivalence point, illustrating the importance of choosing an appropriate indicator.
In general, the faster the pH changes in the range where the indicator changes colour, the sharper the endpoint of the titration will be and the more different indicators will be suitable for the titration.
One limitation of using indicators to find the equivalence point of a titration is that the approximate pH of the equivalence point must be known in order to choose an indicator that will accurately locate the equivalence point. This means that if the approximate pH of the equivalence point is not known, it will be difficult to locate the equivalence point. However,
- Adding a conjugate base to a weak acid
- Adding a strong acid to a weak base
- Adding a strong base to a weak acid
Buffering Capacity and pH
The buffering capacity is the amount of acid or base a buffer can accept without the pH changing appreciably. The greater the amounts of the conjugate acid-base pair, the more resistant they are to change in pH.
If we solve the acid-dissociation-constant expression for [H+] we get:
[H+] = Ka { [HX] / [X-] },
assuming that the amount of added acid or base is less than 5% of the conjugate acid / base molarity.
We can use this to determine the pH of a buffer:
We first take the negative log of both sides:
-log [H+] = -log [Ka { [HX] / [X-] }] = -log Ka - log { [HX] / [X-] }
We know -log [H+] = pH and -log Ka = pKa.
pH = pKa - log { [HX] / [X-] } = pKa + log { [X-] / [HX] }
In general:
pH = pKa + log { [base] / [acid] }
This is known as the Henderson-Hasselbalch equation.
Additions of Acids or Bases to Buffers
Learning Goal 22 Calculate the change in pH of a simple buffer solution, whose composition is known, caused by adding small amounts of strong acid or strong base.
EXAMPLE
A 1 L solution containing 0.100 moles of HC 2 H 3 O 2 and 0.100 moles of C 2 H 3 O 2 forms a buffer solution, pH = 4.74. What is the pH after 0.020 moles of NaOH are added?
STEP 1: Stoichiometry calculation:
The OH-^ will react completely with the weak acid HC 2 H 3 O 2 :
HC 2 H 3 O 2 (aq) + OH-^ (aq) H 2 O (l) + C 2 H 3 O 2 -^ (aq) HC 2 H 3 O 2 OH-^ C 2 H 3 O 2 - INITIAL 0.100 mol 0.020 mol 0.100 mol CHANGE -0.020 mol -0.020 mol +0.020 mol FINAL 0.080 mol 0 mol 0.120 mol
STEP 2: Equilibrium calculation:
The solution contains the weak acid and its conjugate base. We shall therefore use their equilibrium equation, and create an I.C.E. table :
HC 2 H 3 O 2 (aq) H+^ (aq) + C 2 H 3 O 2 -^ (aq) HC 2 H 3 O 2 H+^ C 2 H 3 O 2 - Initial 0.080 M 0 0.120 M Change -x M +x M +x M Equilibrium (0.080 - x) M x M (0.120 + x) M Ka = { [C 2 H 3 O 2 - ][H+] / [HC 2 H 3 O 2 ] } = { (0.120 + x)(x) / (0.080 - x) } = 1.8 x 10- 1.8 x 10-5^ = { (0.120) x / (0.080) } x = [H+] = 1.2 x 10-5^ M pH = -log (1.2 x 10-5) = 4.
Applications of Buffer Calculations
In many systems, such as biological applications, there is extreme sensitivity to minute pH changes. To study these systems, the pH must be controlled by a buffer that is effective within given limits. How does one determine what the buffer concentration is?
Learning Goal 23 Calculate the specific amounts of species necessary to make the buffer effective within given requirements.
EXAMPLE
What must the minimum concentration of HC 2 H 3 O 3 be in a one litre buffer solution of HC 2 H 3 O 3 -C 2 H 3 O 3 -^ (pH = 4.74) if the pH changes by less than 0.1 if 0.050 moles of HCl are added?
STEP 1: Stoichiometry calculation:
The amount of HCl added will react completely with the conjugate base, C 2 H 3 O 3 - :
C 2 H 3 O 3 -^ + H+^ HC 2 H 3 O 3
