Chemistry: Acid-Base Equilibria: Acid-Base Titrations

Seite 1

Acid-Base Titrations, pH Curves and Indicators

Acid-Base Titrations

End Point and Equivalence Point

When you carry out a simple acid-base titration, you usually use an indicator to tell you when the acid and base are mixed in exactly the right proportions to react in equivalent amounts, as dictated by the stoichiometric equation. When the indicator changes colour, this is often described as the "end point" of the titration.

The "equivalence point" is when the acid and base have reacted together in the exact proportions as dictated by the stoichiometric equation. When titrating an aqueous solution of monobasic acid with an aqueous solution of a monoacidic base of the same concentration, 25cm3 of acid will react exactly with 25cm3 of base.

The pH at the equivalence point depends on the combination of acid and base used. For example, if you are titrating aqueous sodium hydroxide with dilute hydrochloric acid then the pH at the equivalence point is 7.00 (at 298K), as both the base and the acid are strong. The solution at equivalence point will contain the salt sodium chloride.

If, however, ethanoic acid (a weak acid) is titrated against sodium hydroxide (a strong base) the solution at the equivalence point will contain the salt sodium ethanoate, and the pH will be greater than 7.00.

If hydrochloric acid (a strong acid) is titrated against aqueous ammonia (a weak base), the solution at equivalence point will contain the salt ammonium chloride, and the pH will be less than 7.00.

Titration of a Strong Acid with a Strong Base

As you add an aqueous solution of an acid to an aqueous solution of a base, you might expect there to be a gradual change in the pH of the solution formed. This is not the case. The shape of the curve when the pH of the solution is plotted against the volume of acid added depends on the nature of the acid and base used. The curves so produced are called "pH titration curves".

We shall start by looking at the pH titration curve when a strong acid is added to a strong base.

The following graph below is produced when adding 1.00moldm-3 HCl(aq) to a 25cm3 sample of 1.00moldm-3 NaOH(aq). The pH is measured using a pH meter. (Graph on the right).

You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge. If you calculate the values, the pH falls all the way from 11.30 when you have added 24.90cm3 to 2.70 when you have added 25.10cm3.

There is a large "steep section" to the curve. As we shall see later in this section, this is an important point to consider when choosing an appropriate acid-base indicator to determine the end point of this titration.

Titration of a Weak Acid with a Strong Base

For this example we shall add 1.00moldm-3 ethanoic acid to 25cm3 of 1.00moldm-3 sodium hydroxide. (Graph on the left).

The curve is the same as that for a strong acid-strong base up to the equivalence point, but there is a difference once the acid is present in excess. Past the equivalence point the solution contains a mixture of ethanoic acid and sodium ethanoate. This mixture acts as a buffer solution and hence resists any large change in pH upon addition of further acid.

Titration of a Strong Acid with a Weak Base

For this example we shall add 1.00moldm-3 hydrochloric acid to 25cm3 of 1.00moldm-3 aqueous ammonia. (See graph above right).

When the acid is first added the pH starts to fall quite sharply, but the curve quickly levels out. This is ammonia and ammonium chloride. Notice the pH at the equivalence point is less than 7 because the salt formed, ammonium chloride, is composed of a strong acid and a weak base.

Titration of a Weak Acid with a Weak Base

For this example we shall add 1.00moldm-3 ethanoic acid to 25cm3 of 1.00moldm-3 aqueous ammonia.

Notice that there is not any steep section to this graph. Instead, there is what is known as a "point of inflexion". The lack of a steep section means that it is difficult to do a titration of a weak acid against a weak base using an indicator.

Acid-Base Indicators

Earlier in this section, we looked at pH titration curves obtained when an aqueous solution of an acid is added to an aqueous solution of a base. The four cases we considered were:

1. Strong acid-strong base

2. Weak acid-strong base

3. Strong acid-weak base

4. Weak acid-weak base

We are now going to use there four curves to help us understand why different indicators are required for difference types of titrations.

An acid-base indicator is either a weak acid or a weak base. Most indicators are weak acids (HIn). For an indicator that is a weak acid, its dissociated in aqueous solution can be represented as:

HIn(aq) -><- H+(aq) + In-(aq)

The molecule, HIn and its conjugate base, In- have different colours in aqueous solution. For methyl orange these are red and yellow respectively.

HIn(aq) -><- H+(aq) + In-(aq)

Red Yellow

When [H+(aq)] is sufficiently large the equilibrium will shift far enough to the left for the red colour to predominate. If [H+(aq)] is very low then the equilibrium will lie far over to the right and the yellow colour will predominate. Hence, the indicator changes colour according to the pH of the solution.

There will be a stage at which [HIn(aq)] = [In-(aq)], and the indicator will appear orange. The exact pH at which this stage is reached can be determined using the equilibrium constant, KIn, for methyl orange.

KIn = [H+(aq)][In-(aq)] / [HIn(aq)] = 2.00 x 10^-4 moldm-3.

When [HIn(aq)] = [In-(aq)], the expression becomes:

[H+(aq)] = 2.00 x 10^-4moldm-3.

This gives a pH of 3.70 for the "half-way" stage. So, methyl orange will change colour at a pH of 3.70. Note this pH value is also the same as the value of pKIn for methyl orange. Therefore, the pH at which different indicators change colour can be determined from their pKIn values.

pH Range of Indicators

As a "rule of thumb", the red colour of methyl orange will first predominate when [HIn(aq)] is ten times [In-(aq)], and the yellow colour will predominate when [In-(aq)] is ten times [HIn(aq)].

The approximate pH at which each colour predominates can be calculated as follows.

When [HIn(aq)] = 10[In-(aq)]:

[H+(aq)][In-(aq)] / 10[In-(aq)] = 2.00 x 10^-4 moldm-3.

[H+(aq)] = 2.00 x 10^-3 moldm-3.

So, the pH at which the red colour first predominates is 2.70.

A similar calculation will show that the pH at which the yellow colour first predominates is 4.70. The pH range of methyl orange is therefore approximately 2.70 to 4.70.

The exact pH range of methyl orange is 3.10 to 4.40. This means that at pH below 3.10, methyl orange will appear red. At pH above 4.40, methyl orange will appear yellow. Between 3.10 and 4.40, methyl orange will be a shade of orange.

Choice of Indicator

A good indicator shows a complete colour change upon the addition of one drop of acid from the burette. This is necessary in order to accurately determine the end point of the titration.

Therefore, there has to be a minimum pH change equivalent to the pH range of the indicator in order for the indicator to successfully determine the end point.

Strong acid-strong base titration

Let us first consider the use of methyl orange and phenolphthalein as indicators for a strong acid-strong base titration.

The diagram below shows the pH titration curve for 25cm3 of 1.00moldm-3 NaOH(aq) titrated with 1.00moldm-3 HCl(aq). The pH changes of methyl orange and phenolphthalein have also been included.

You will notice that the pH range of each indicator falls within the steep section of the curve, where a large pH change is occurring upon the addition of just one drop of acid. This means that both indicators will change colour at the end point and therefore both are suitable indicators to use in this titration.

Weak Acid-Strong Base Titration

We shall now consider the suitability of each indicator for a weak acid-strong base titration.

The graph below shows the pH titration curve for 25cm3 of 1.00moldm-3 NaOH(aq) titrated with 1.00moldm-3 CH3COOH(aq). Once again, the pH ranges of methyl orange and phenolphthalein have also been included.

This time, only the pH range of phenolphthalein falls within the steep section of the curve. So, phenolphthalein is suitable, but methyl orange is not.

Strong Acid-Weak Base Titration

The graph below shows the pH titration curve for 25cm3 of 1.00moldm-3 NH3(aq) titrated with 1.00moldm-3 HCl(aq). Again, the pH ranges of methyl orange and phenolphthalein have also been included.

This time, only the pH range for methyl orange falls within the steep section of the curve. So, methyl orange is suitable, but phenolphthalein is not.

Weak Acid-Weak Base Titration

The graph below shows the pH titration curve for 25cm3 of 1.00moldm-3 NH3(aq) titrated with 1.00moldm-3 CH3COOH(aq). Again, the pH ranges of methyl orange and phenolphthalein have also been included.

As there is no prominent steep section of the curve, neither indicator is suitable. In fact, the end point of a titration of a weak acid and a strong base cannot be determined using an acid-base indicator. The end point of such a titration is best determined by measuring the temperature changes (thermometric titration) or electrical conductivity charges (conductometric titration).

Choosing the Best Indicator

The best indicator to choose for a particular titration is the one whose kPa value is as close as possible to the pH at the equivalence point.

Bromothymol blue (pKIn = 7.00) is a particularly good indicator for a strong acid-strong base titration.

Seite 2

Buffer Solutions

What is a Buffer Solution?

Many experiments, particularly in biochemistry, have to be performed in aqueous solutions of fairly constant pH. Unfortunately, it is impossible to make a solution whose pH is totally unaffected by the addition of even small amounts of acid or base. However, it is possible to make a solution whose pH remains almost unchanged when small amounts of acid or base are added. Such a solution is called a buffer solution.

There are many ways of making a buffer solution but two of the most common are:

1. Mix a weak acid with its conjugate base.

2. Mix a weak base with its conjugate acid.

A Buffer Made from a Weak Acid and its Conjugate Base

The simplest example of this is ethanoic acid and sodium ethanoate. The salt of the weak acid has to be soluble in water, which is why sodium and potassium salts are commonly used to make buffer solutions.

In this mixture the acid is partially dissociated, whereas the salt is fully dissociated. The relevant equations are:

CH3COOH(aq) -><- CH3COO-(aq) + H+(aq)

CH3COONa(aq) -> CH3COO-(aq) + Na+(aq)

This mixture will produce a buffer solution with a pH value less than 7. The exact pH depends on the concentration of both acid and its conjugate base, and can be calculated as follows.

Calculating the pH of a Buffer Solution

For our example we shall use a solution that has been made by mixing equal volumes of 1.00moldm-3 ethanoic acid and 1.00moldm-3 sodium ethanoate at 298K.

If we assume that the extent of dissociation of the acid is negligible, then the concentration of CH3COOH at equilibrium will be 0.500moldm-3.

Also, again because the dissociation of the acid is negligible, the concentration of the ethanoate ions at equilibrium can be considered to be entirely made up from the sodium salt. So, the concentration of ethanoate ions at equilibrium is also 0.500moldm-3.

Note that the concentration of both acid and salt in the mixture is half of the concentrations used because equal volumes of each solution were mixed. The total volume of the mixture is twice that of the original volume of each solution used.

At 298K,

Ka = [CH3COO-(aq)][H+(aq)] / [CH2COOH(aq)] = 1.74 x 10^-5 moldm-3.

Rearranging this equation and substituting values for [CH3COOH(aq)] and [CH3COO-(aq)] gives:

[H+(aq)] = (1.74 x 10^-5 x 0.500) / 0.500 moldm-3

= 1.74 x 10^-5 moldm-3.

This gives a pH for the buffer solution of 4.76.

How does the Buffer Action Work?

When a small amount of acid is added to the buffer solution, the majority of the H+ ions added react with the CH3COO- ions to form CH3COOH molecules:

CH3COO-(aq) + H+(aq) -> CH3COOH(aq)

When a little base is added, the majority of the OH- ions added react with the CH3COOH molecules:

CH3COOH(aq) + OH-(aq) -> CH3COO-(aq) + H2O(l)

A new equilibrium mixture will be established in which the concentration of both CH3COOH and CH3COO- will have changed slightly from their original values. So, there will be a change in pH, but this will be minimal. This is shown by the following argument.

To show that the pH has changed very little, we once again need to make use of the expression for Ka for the acid:

[CH3COO-(aq)][H+(aq)] / [CH2COOH(aq)] = 1.74 x 10^-5 moldm-3.

Rearranging this equation gives:

[CH3COO-(aq)] x Ka / [CH2COOH(aq)] = [H+(aq)]

As there are a relative large number of CH3COOH molecules in the solution, as the extent of dissociation of the acid is very small, the change in [CH3COOH(aq)] will be negligible.

Also, as there are a relatively large number of CH3COO- ions present, resulting from the total dissociation of the CH3COONa, the change in [CH3COO-(aq) will be negligible.

This means that the ration [CH3COOH(aq)] / [[CH3COO-(aq)] will remain fairly constant.

If this ratio remains fairly constant, then [H+(aq)] remains fairly constant, because at a given temperature, Ka is also constant.

If [H+(aq)] remains fairly constant, then the pH remains fairly constant.

To summarise, a buffer solution of a weak acid and its conjugate base maintains a fairly constant pH because the ratio of [CH3COOH(aq)] to [CH3COO-] remains fairly constant when small amounts of either acid or base are added.

The best way to demonstrate the effect of adding a small amount of acid on the pH of a buffer solution is to calculate the pH of the buffer both before and after adding the acid.

We have already calculated the pH of a buffer solution made by mixing equal volumes of 1.00moldm-3 ethanoic acid and 1.00moldm-3 sodium ethanoate at 298K. It is 4.76.

Let us imagine that we have mixed 500cm3 of each solution to make 1dm3 of buffer solution. The amount CH3COOH and CH3COO- in this solution will both be equal to 0.500mol.

Let us imagine we have 1dm3 of this solution and we add 1.00 x 10^-2 (0.0100) mol of HCl to it. To make the mathematics easier, we shall assume that the volume of the solution does not change.

0.0100 mol of HCl will provide 0.0100 mol of H+ ions. These will react with the CH3COO- ions in the buffer in a 1:1 molar ratio:

CH3COO-(aq) + H+(aq) -> CH3COOH(aq)

The amount of CH3COOH present will now have increased from 0.500 to 0.510mol.

The amount of CH3COO- present will have decreased from 0.500 to 0.490mol.

So, the new concentrations of acids and base present are:

[CH3COOH(aq)] = 0.510moldm-3

[CH3COO-(aq)] = 0.490moldm-3

The new hydrogen ion concentration is therefore, 1.81 x 10^-5moldm-3.

The pH is now 4.74.

The pH has changed by 0.02 units from 7.76 to 7.74.

To appreciate how effective the buffer solution is in controlling the pH, let us consider adding 0.0100mol of H+ ions to 1dm3 of deionised water.

The pH, at 289K, of deionised water is 7.00 so:

[H+(aq)] = 1.00 x 10^-7 moldm-3/

Adding 0.0100mol of H+ ions gives:

[H+(aq)] = (1.00 x 10^-7 + 0.0100) = 0.0000001moldm-3

The new pH is 2.00

The pH has dropped by 5 units as opposite to 0.02 units with the buffer solution. A considerable difference.

The Henderson-Hasselbalch Equation

If you are convinced that you can remember equations, then the Henderson-Hasselbalch equation can be used to calculate the pH of a buffer solution.

For a weak acid, HA, and its conjugate base, A-, the following equation applies:

[H+(aq)] = Ka x [HA(aq)] / [A-(aq)]

The concentration terms in the equation are, technically speaking, the concentrations at equilibrium. However, for reasons we have already discussed, it is reasonable to take the original concentrations.

The original concentration of A- will be the same as that of the salt, providing a sodium or potassium salt has been used. The equation can now be written as:

[H+(aq)] = Ka x [acid] / [salt]

If we get the logarithm to base ten of both sides of this equation we get:

lg[H+(aq)] = lgKa + lg[acid]/[salt]

or

pH = pKa + lg[acid]/[salt]

This last equation is the most common form of the Henderon-Hasselbalch equation.

If we use the Henderson-Hasselbalch equation to calculate the pH of our solution containing ethanoic and ethanoate ions, both of concentration 0.500moldm-3 we obtain:

pH = 4.76 + lg(0.500/0.500) = 4.76

The same answer we obtained earlier!!!!!!!!!!!!!! why is the damn textbook doing this to meeeeeeeee???????????] just another stupid equation they think they can chuck in when I'm tired and this is a firggin long topic. wt actual f.

A Buffer Made from a Weak Base and its Conjugate Acid

Here we go the same bloody thing again with lots of equations I have to type and which take foreeeever. guys this is a lesson not to work for like 6 hours straight. ok that's exaggerating but its definitely been at least 4. sorry this isn't the actual topic....

The most common example of this type of buffer is ammonia and the ammonium ion. The ammonium ion is usually supplied in the form ammonium chloride. This mixture will provide a buffer solution with a pH greater than 7.

The most convenient equilibrium to consider is:

NH4+(aq) -><- NH3(aq) + H+(aq)

This mixture provides a relatively high concentration of both NH3 molecules and NH4+ ions.

The buffer works in a similar manner to the weak acid-conjugate base system and the addition of small amounts of either acid or base. This results in the pH remaining fairly constant because it is given by the equation:

pH = pKa + lg[NH3(aq)] / [NH4+(aq)]

As pKa is constant at a given temperature, the pH of the solution depends on the ratio of [NH3(aq)] to [NH4+(aq)].

How to Make a Buffer Solution with a Required pH

To make a buffer solution with a pH less than 7, you need to use a mixture of a weak acid and its conjugate base.

Conversely, to make a buffer solution with a pH greater than 7, you need to use a mixture of a weak base and its conjugate acid.

Controlling the pH of Blood

The human body operates within a narrow range of pH values. For example, the pH of arterial blood plasma needs tobe in the range of 7.35 to 7.45. If the pH of this blood plasma were to change significantly, particularly if it were to fall, the way that the whole body functions would be affected.

The pH of blood is controlled by a mixture of buffers, the most important of which is the carbonic acid-hydrogencarbonate buffer mixture.

In this mixture the carbonic acid molecule, H2CO3, acts as the weak acid. The hydrogencarbonate ion, (HCO3)-, is the conjugate base of H2CO3.

The equilibrium that exists is represented by the equation:

H2CO3(aq) -><- (HCO3)-(aq) + H+(aq)

Under normal circumstances the amount of (HCO3)- ion present is approximately 20 times that of H2CO3. As normal metabolism produces more acids than bases, this is consistent with the needs of the body.

Any increase in the concentration of hydrogen ions in the blood, by for example the production of lactic acid in the muscles, results in the equilibrium shown above moving to the left as the added H+ ions react with the (HCO3)- ions.

The pH of the blood can also be raised by a variety of respiratory and metabolic causes. For example, the overuse of diuretics increases the amount of urine excreted from the body. If the urine contains large amounts of acids then the pH of the blood will increase. If this happens, the equilibrium shown above will move to the right as the H2CO3 molecules ionise to increase the H+ concentration and restore the pH to its normal level.

This is considered to be the most important buffer because it is coupled with the respiratory system of the body. Carbonic acid is not particularly stable and in aqueous solution it decomposes to form carbon dioxide and water.

H2CO3(aq) -><- CO2(aq) + H2O(aq)

It is the respiratory system that is responsible for removing carbon dioxide from the body. Aqueous carbon dioxide exists in equilibrium with gaseous carbon dioxide:

CO2(aq) -><- CO2(g)

Combining these reactions gives:

CO2(g) + H2O(aq) -><- CO2(g) + H2O(l) -><- H2CO3(aq) -><- (HCO3)-(aq) + H+(aq)

When these equilibria shift to the left as a result of an increase in hydrogen ion concentration, the concentration of carbon dioxide in the blood increase. The carbon dioxide leaves the blood in the lungs and is then exhaled, thus maintaining the normal pH of the blood.

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Buffer Solutions and pH Curves

Buffer Action During a Titration

The diagram shows a typical pH curve obtained when a weak acid is titrated against a strong base.

The region of relatively flat curve at approximately pH = 4 is known as the buffer range. This is the change in pH as the base is added and is gradual. Over this range there is a considerable concentration of both acid and conjugate base molecules. This mixture is displaying buffer action.

Determining Ka from a pH Titration Curve

This experiment involves performing a titration with an aqueous solution of a weak acid in the conical flask. A standard solution of a strong base, such as sodium hydroxide, is added from a bottle and the pH of the solution is measured after each addition.

A graph of pH against volume of base added is then plotted. From the graph, the minimum volume of base required to completely react with all of the acid is determined. This is the volume at the equivalence point.

The graph is then used to determine the pH at the half-equivalence point. This pH value is equal to the pKa value of the weak acid. From this it is a simple matter to calculate Ka for the acid.

The volume at the equivalence point is 25cm3, so the volume at the half-equivalence point is 12.5cm3. The pH when 12.5cm3 of base is added is 4.80.

The pKa of the acid = 4.80

Ka for the acid = 1.58 x 10^-5 moldm-3.

Supporting Theory

The theoretical justification for determining Ka for a weak acid by the method shown above is quite straightforward.

The mixture at the half-equivalence point is a buffer solution (as stated above). The pH of a buffer solution is calculated using the following equation:

pH = pKa + lg[salt] / [acid]

At the half-equivalence point:

[salt] = [acid], so [salt] / [acid] = 1.

The logarithm to the base 10 of 1 (lg(1)) = 0.

So the equation becomes:

pH = pKa

Alternative Method

This method has been called the "half-volume method".

1. Using a volumetric pipette, place 25.0cm3 of an aqueous solution of the weak acid into a conical flask.

2. Add a few drops of phenolphthalein indicator.

3. Titrate against a solution of aqueous sodium hydroxide until the end point colour is obtained.

4. Note the volume of sodium hydroxide required. This is the minimum volume required to completely react with the acid.

5. Use a fresh 25.0cm3 sample of the same aqueous solution of the weak acid and the same aqueous solution of sodium hydroxide, but this time do not add the phenolphthalein.

6. Add only half the volume of sodium hydroxide required to react with the acid.

7. Measure the pH of this solution. This pH value is equal to the pKa value of the acid.

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Enthalpy Changes of Neutralisation for Strong and Weak Acids

Standard Enthalpy Change of Neutralisation

In the first year, we defined the standard enthalpy change of neutralisation as follows:

The standard enthalpy change of neutralisation is the enthalpy change measured at 100kPa and a specified temperature, usually 298K, when one mole of water is produced by the neutralisation of an acid with an alkali.

We also explained that the standard enthalpy change of neutralisation of a strong acid with a strong base will always have a very similar value, because strong acids and strong bases are almost fully dissociated in aqueous solution. This value is around -57.6 kJmol-1.

It is interesting to note that for two of the weak acids, CH3COOH and HCN, the magnitude of the value of enthalpy change of neutralisation is less than for a strong acid-strong base, but the magnitude of the value for HF is greater than 57.6

The equation for the reaction between a weak acid, HA, and a strong base such as sodium hydroxide can be represented as:

HA(aq) + OH-(aq) -> A-(aq) + H2O(l)

The major difference in this reaction, compared to that of a strong acid with a strong base, is that the majority of the acid molecules are undissociated. Obviously energy will be required to dissociate them, and this may explain why less heat energy is given out when both CH3COOH and HCN are neutralised by NaOH.

However, it does not explain why more heat energy is given out when HF is neutralised by NaOH. There must be at least one other energy term involved.

In the above argument, we have ignored the fact that, once the acid molecules are dissociated (which is an endothermic process), the ions formed will then be hydrated by the water molecules present in the aqueous solution. Hydration of ions is an exothermic process, so heat energy will be guaranteed. The two processes can be represented by the following equations for a weak acid of formula HA#:

HA(aq) -> H+(aq) + A-(aq) DISSOCIATION (endothermic) xkJmol-1

H+(g) + A-(g) -> H+(aq) + A-(aq) HYDRATION (exothermic) ykJmol-1

If the magnitude of x is greater than that of y, then the overall heat energy change involved in dissociating the acid molecules and then hydrating the ions will be endothermic. This is the case with CH3COOH and HCN.

However, with HF, the enthalpy change of hydration of the ions releases more heat energy than the dissociation of the molecules takes in. Hence the overall effect is that the standard enthalpy change of neutralisation of HF by NaOH is greater than 57.6kJmol-1.