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Unifying Concepts

Bringing back quite a bit of How Far, How Fast?, it might be good revision to have a quick look through that topic before moving to Unifying Concepts.

Rate Equations

You should already know that:

We can calculate this using a special type of equation called a rate equation. Rate equations are related to chemical equations. If we had the chemical equation: mA + nB ----> products, the rate equation would look like this:

rate = k[A]m[B]n mol dm-3s-1

  • Every rate equation has a rate constant, k. The larger the value of k, the faster the equation goes.
  • If the temperature of a reaction is increased, we know the rate increases due to the collision theory. This also affects the rate constant (k), making it bigger, as the overall rate increases. The vice versa is true: lower the temp, and the k decreases as the rate decreases.
  • The reactants' initial concentrations are A and B.
  • These initial concentrations are taken to the power of the number of moles in the chemical equation (m and n).

For example, if we had I2 + 2S2O32- -----> 2I- + S4O62-, the rate equation would be:

rate = k [I2][S2O32-]2 mol dm-3s-1

Because there is one mole of I2, we call it first order with respect to I2 and because there are two moles of S2O32-, we call it second order with respect to S2O32-.

The total order of reaction, over all, is just a case of adding together the individual orders. 1+2=3. This reaction is third order overall.

So, if we have k=100 mol-2dm6s-1, initial concentration of 0.03 mol dm-3 of I2 and 0.05 mol dm dm-3 of S4O62-, the rate would be:

rate = 100 x 0.03 x 0.052
rate = 3 x 0.0025
rate = 0.0075 mol dm-3s-1

If we rearrange the rate equation, we can calculate k quite easily.

The units are more interesting. Rate, of course, is mol dm-3s-1 and each concentration is mol dm-3.

However, if we have [A][B]2 for example, we'd have mol dm-3 and another the mol dm-3 is squared to get mol2dm-6.

The mol dm-3s-1 is divided by the bottom bit:

The two mol dm-3 cancel out, leaving us:

Using basic rules of indices, we can move the bottom part to the top, to get:

s-1 mol-2dm6

Not too hard, huh?

A question:

KI and K2S2O8 react in the following equation:

KI(aq) + K2S2O8(aq) -----> 2K2SO4(aq) + I2(aq)

The initial concentration of KI is 1.0 x 10-2 mol dm-3 and the initial concentration of K2S2O8 is 5.0 x 10-4 mol dm-3. The initial rate of disappearance of K2S2O8 is 1.02 x 10-8 mol dm-3. Construct a rate equation and find the rate constant.

The rate equation uses the products, in the above equation, and so is:

rate = k [KI(aq)] [K2S2O8(aq)]

In order to find k, we can rearrange this to make k the subject of the formula:

Now it's just a case of putting in the numbers to calculate k:

k = 2.04 x 10-4

Next, we figure out the units:

The mol dm-3 on the top and the bottom cancel, leaving: mol-1dm3s-1.

So, put together, we get the answer:

k = 2.04 x 10-4 mol-1dm3s-1.

Rate mechanism

We've encountered various mechanisms over time. A rate mechanism is a description of a chemical reaction as a series of one-step processes.

Some of these steps might be slower than others. The slowest step is known as the rate determining step because, as the reaction cannot go faster than the slowest step, it determines the rate of the reaction.

How do we know what the rate determining step is? Through something called the rate law. This law states that the order of reaction with respect to a reactant indicates how many molecules of that reactant are involved in the rate determining step.

For example, if the reaction is first order with respect to a reactant, there will be one mole of that reactant in the rate determining step. If it's second order, there will be two moles.

If we take the following:

H2(g) + 2ICl(g) -----> I2(g) + 2HCl(g)

Rate: k [H2][ICl]

In the stoichiometric equation (fancy name for the equations we usually see), we see one mole of H2 reacts with two moles of ICl(g). However, if we look at the rate equation, we see that one mole of H2 reacts with only one mole of ICl. This means there must be a slower step in the overall equation that reacts with a ratio of 1:1.

If the number of moles in the overall equation don't correspond to those in the rate equation, you know that there is a RDS step somewhere. From the above example, it could be:

H2 + ICl -----> HCl + HI

 

Thanks to Benjamin for the following extra info on Rate Mechanisms:

When we write a chemical equation such as:

2 C2H2 + 5 O2 → 4 CO2 + 2 H2O

we see that two molecules of C2H2 react with 5 molecules of oxygen. According to the collision theory, molecules must collide in order to react. But it is highly unlikely that 7 molecules would collide together all at once.

Instead, the reaction most likely occurs in a series of simple steps which only required two or three molecules colliding at any one instant. Although these steps cannot always actually be observed, chemists can often make predictions about the sequence of events.

For example, nitrogen monoxide reacts with oxygen according to the equation

2 NO(g) + O2 → 2 NO2

This reaction does not occur in a single step, however, but rather through these two steps:

Step 1: 2 NO → N2O2
Step 2: N2O2 + O2 → 2 NO2

Notice that if you add these two reactions together, you end up with the overall reaction:

Step 1: 2 NO → N2O2

Step 2: N2O2 + O2 → 2 NO2

Overall reaction:

2 NO(g) + O2 → 2 NO2

The series of steps a reaction undergoes is called the reaction mechanism.

If you have any information to add, please email it!

Concentrations

It's possible to measure the concentration of reactants as a reaction occurs. There are two different methods, measuring the rate against the concentration and the concentration against the time. They produce quite different graphs, however, so you need to be clear of them in your head. The graphs are different, too, depending on whether the reactions are zero, first or second order, so are useful for distinction.

Zero order

The rate equation is rate = k[A]0
The gradient is constant, here, because the rate of reaction is constant, irrespective of [A].

First order

The rate equation is rate = k[A]1

Though this graph looks similar to the second order one below, you can tell it is first order because the half lives (t½) are constant. Half life is the time it takes for half of the concentration to disappear. If you measure the concentration at 0.1 mol dm-3, for example, the half life will be the time taken to reach 0.05 mol dm-3.

 
Second order The rate equation is rate = k[A]2

The t½ is not constant.

Considering both first and second reaction, we see a steep gradient at the beginning for a fast rate of reaction and a shallow gradient near the bottom, showing a slow RoR.

Half life can greatly vary, depending on reactions, from seconds, to minutes, to hours, to years. In order to calculate it on an exponential, use the following method:

  1. Choose a convenient concentration that can be easily halved along the y-axis.
  2. Draw a horizontal line to the graph and a vertical line down to find the x value.
  3. Half the y-value, and draw a line from this point to the graph, and then another vertical line to the x-axis, to find the next x-value.
  4. Find the difference between the two x-values and this is the half life.
  5. Repeat the steps at a different initial concentration. If the half life is the same, then it's first order; if not, it's second order.
  6. Find a third half life just to be certain, however!

The gradient of any concentration-time graph gives us the rate of reaction. If a question asks us to find the rate of reaction at a particular point, we need to draw a triangle from the point on the graph:

If we divide the height of this triangle by its width, we get the rate of reaction... easy!

Initial rate method

Another way of figuring out whether reactants in an equation are first, second or zero order is to use the initial rate method. This means that the initial rate is found in several experiments, using different concentrations. It's usual to just change one concentration per equation, and see how this affects the rate of reaction.

  • If a reactant is zero order, no matter what concentration it is, the rate of reaction stays the same.
  • If a reactant is first order, the rate will increase proportionally as the concentration increases.
  • If the reactant is second order, whatever the value the concentration increases by, the rate will increase by the square of this.

So, to take O2(g) + 2ClO(g) ------> 2ClO2(g) you might be presented by a table like this:

Experiment Number Initial concentration of O2 (in mol dm-3).

Initial concentration of ClO (in mol dm-3).

Initial rate of disappearance of ClO.
1 1.0 1.0 0.7
2 1.0 2.0 1.4
3 2.0 2.0 5.6
4 6.0 6.0 151.2

 

Now, if you look at experiments 1 and 2, you'll see that ClO concentration is being increased and O2 is staying constant. ClO goes from 1.0 to 2.0, and so is doubled. If we look at the rate of disappearance, we see that it goes from 0.7 to 1.4, and so is also being doubled. The rate increases proportionally to the concentration, so the order of reaction for ClO is first order.

If we consider experiment 2 and 3, we see that the O2 concentration is increased, while the ClO stays constant. The O2 goes from 1.0 to 2.0, and so is doubled. The rate goes from 1.4 to 5.6, and so is multiplied by four...or 22. This means that it is second order, because the rate change is the square of the concentration change.

So, the overall rate equation will look like this:

rate = k [ClO(g)]1[O2(g)]2

We can find k by the usual process:

k = 0.7 (1 x 12)

k = 0.7 mol-2 dm6 s-1

That's just about all you need to know for this section... get learning!

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Kc

Now we've looked at the How Fast part of this topic, it's time to consider the How Far section. Everything in this section refers to equibriums and equilibrium positions.

Kc is something called the equilibrium constant.

It's easiest to explain, if we look at it in terms of an equilibrium's general equation: aA(aq) + bB(aq) cC(aq) + dD(aq). From this, the equilibrium law states that:

In other words, Kc are the products divided by the reactants.

  • A large Kc means that the equilibrium is to the RHS, as lots of products and few reactants are being formed...i.e. a large product yield.
  • A small Kc means the equilibrium is to the LHS, as lots of reactants and few products are being formed...i.e. a low product yield.

The units for Kc vary, depending on the equilibrium in question. Each part is in mol dm-3, but if the reactant/product is squared due to two moles, for example, the part becomes mol2 dm-6.

Take the following equilibrium, for example:

2SO2(g) + O2(g) -----> 2SO3(g)

The Kc here would be:

and the units would be:

A question might give you some volumes of the gases and ask you to work out the value for Kc.

For example:

Calculate the Kc for

H2(g) + CO2(g) CO(g) + H2O(g)

given that [H2(g)]eq = 0.53 mol dm-3, [CO2(g)]eq = 80.53 mol dm-3, [CO(g)]eq = 9.47 mol dm-3and [H2O(g)]eq = 9.47 mol dm-3.

The Kc equation is:

Sticking in the numbers, we get:

Times and divide these numbers and we get an answer of 2.1. This has no units because the mol dm-3s on the top are cancelled out by those on the bottom.

If you've given initial concentrations rather than equilibrium concentrations, you need to figure out what the equilibrium concentrations are before you can work out the Kc.

For example:

0.067 mol of SO3 is placed in a equilibrium vessel (volume of 1.52 dm3) and allowed to reach the following equilibrium.

2SO3(g) 2SO2(g) + O2(g)

At equilibrium, the concentration of SO2 is 0.00142. Calculate Kc for the equilibrium.

First off, draw up a table of what you have and fill in what you know:

  2SO3(g) 2SO2(g) O2(g)
Initial amount in mol. 0.067 0 0
Equilibrium amount in mol.   0.00142  
Equilibrium concentration in mol dm-3      

Next, work out how many moles of SO3 have been used up. The reaction is 2mol---->2mol, so 0.00142 must have been used. Take this away from the initial concentration to get 0.05668 mol.

Then figure out how many moles of O2 have been produced. The reaction in this case is 2mol---->1 mol, so divide 0.00142 by 2 to get 0.00071 mol.

  2SO3(g) 2SO2(g) O2(g)
Initial amount in mol. 0.067 0 0
Equilibrium amount in mol. 0.05668 0.00142 0.00071
Equilibrium concentration in mol dm-3      

In the original question we had a volume of 1.52 dm3, so divide the moles by this to get the equilibrium concentrations:

  2SO3(g) 2SO2(g) O2(g)
Initial amount in mol. 0.067 0 0
Equilibrium amount in mol. 0.05668 0.00142 0.00071
Equilibrium concentration in mol dm-3 0.0431 0.000934 0.000467

These final figures are the ones we use in the Kc equation:

Multiply these through and work out the units and you get the answer:

2.2 x 10-7 mol dm-3

Conditions

Kc only changes if the temperature changes. Although the equilibrium position may change with pressure, resulting in more or less yield, the Kc remains constant... (if you could explain this to me, I'd appreciate it!)

So, what happens to Kc when the temperature changes? It depends on the equilibrium in question. If we think back to AS How Far, How Fast, we might recall Le Chatalier's Principal. Chatalier said that if there is a change to an equilibrium, the equilibrium will change position in order to oppose the change.

If an equilibrium is exothermic in the forward direction and we increase the temperature, the reaction will move to the endothermic LHS, producing lots of reactants. Therefore, the Kc decreases.

The reverse is true: if the equilibrium is exothermic in the forward direction and we decrease the temperature, the equilibrium will move to the RHS, with a high yield, and so Kc will increase.

If, instead, an equilibrium is endothermic in the forward direction, an increase in temp will move it to the RHS with a higher yield, and so Kc will increase.

If an equilibrium is endothermic in the forward direction, a decrease in temperature will move it to the LHS, decreasing the yield, and so decreasing Kc.

Not too bad, hmm? Just remember that pressure doesn't affect it!

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Kp

Kc is a measure of equilibrium in aqueous solutions. However, for equilibriums that involve gases, it's often much more convenient to measure pressure rather than concentration. Instead of Kc, we use something called Kp, which means expressing the equilibrium in terms of partial pressures.

Each gas has a partial pressure, which is the pressure exerted on that gas alone in the same volume as the mix of gases. The total pressure of the gas mixture (ptotal) is the sum of all the partial pressures added together. ptotal = p1 + p2 + p3 + ....

To work out the partial pressure on a gas in the equilibrium, we use the following equation:

pA = mole fraction x ptotal

The mole fraction is the number of moles of that individual gas, divided by the total moles in the equation.

As with Kc, we have the following equation for Kp:

The partial pressures of the products (C and D) go on the top of the equation and the partial pressures of the reactants (A and B) on bottom. The number of moles of A, B, C, and D determine to what power they are raised.

For example, the Kp of the equilibrium PCl3(g) + Cl2(g) PCl5(g) would be:

and the units would be kPa-1.

So a question might look something like this:

In the equilibrium N2O4(g) 2NO2(g), there were 0.1 mol of N2O4 and 0.4 mol of NO2. The total pressure was 100 kPa. Work out the partial pressures of each, and the Kp.

Okay, so first off, we work out the mole fractions for both. There are a total of 0.1 + 0.4 mol of gas, so 0.5 mol. We divide each number of moles by this to get their mol fraction, and then times this by the total pressure:

p(N2O4) = 0.1 / 0.5 = 0.2 x 100 kPa = 20 kPa. p(NO2) = 0.4 / 0.5 = 0.8 x 100 = 80 kPa.

Next, we add these into the kPa equation:

kp = 320

Finally, work out the units... (kPa)(kPa) / (kPa)... so one on the top and bottom will cancel, leaving kPa.

Nothing to it!


Acids and Bases

This is quite an important topic, particularly for any biologists out there. If you think back to How Far, How Fast, you'll probably remember the Bronsted-Lowry Theory of acids and bases:

  • Acids are proton (H+) donors.
  • Bases are proton acceptors.
  • In an acid-base reaction, the acid donates a proton to form a conjugate base, while the base accepts a proton to form a conjugate base.

For example:

HCl + H2O H3O+ + Cl-
acid + base   conjugate
acid
+ conjugate
base

Bases must always have at least one lone pair of electrons, in order to accept the H+ ion and form a dative covalent bond.

In solutions, H+ can't really exist, because it is immediately attracted to the lone pairs on the oxygens in water, to form the H3O+ ion. This is known as the oxonium ion or the hydroxonium ion.

Water can act as an acid if a strong enough base is present, for example:

H2O + NH3 NH4+ + OH-
acid + base   conjugate
acid
conjugate
base

Because it can act as either acid or base, water is said to be amphoteric.

Strong Acids

Acid strength depends on how much an acid dissociates. For strong acids, we assume this is 100%. Weak acids only partially dissociate. The more the dissociation, the stronger the acid. In a reaction between two 'acids', the stronger will force the other to act as a base, such as in a nitric-sulphuric equilibrium:

HNO3 + H2SO4 H3SO4+ + NO3-
acid + base   conjugate
acid
conjugate
base

Why are some acids stronger than others? It's not enough to say that one is a stronger proton donor, and depending on the acid in question, there are two possible answers.

When comparing phenol's acidity to ethanol's, there are two reasons why phenol is more acidic. First, the bond between the proton and the rest of the molecule is weakened. The electrons in the bond between O-H are delocalised into the benzene ring, thus weakening the bond. Also, the anion is stable, as the charge delocalised, meaning it is more likely that it forms.

Between HF and HCl, the strongest acid is HCl. The bond between H and F is very strong, meaning it is less likely to dissociate into its ions.

Measuring pH

Because the [H+] concentration can vary widely, it's impractical to measure acid strength on concentration. Instead, we use the concentration in a log calculation, to result in an answer in a much smaller range (usually, pH=1-14, but be aware that negative pHs can be found, and alkalis stronger than pH=14.)

In order to measure the pH of a strong acid, we must assume that the acid fully dissociates.

We use the concentration of the acid in the following calculation (which you must learn!):

pH = - log10[H+]

If you're ever asked to find the [H+(aq)] concentration from the pH, you need to use the reverse calculation:

[H+(aq)] = 10-pH

So, if you have 0.01 mol dm-3 HCl, simply put it into the log calculation:

pH = - log10 0.01

pH = 2

Pretty straightforward, hmm?

To make things more tricky, the exam question might say something like this:

25cm3 of 0.1 mol dm-3 HCl was made up to 100 cm3 with water. What was the pH of the resulting solution?

Here, we need to use the mol calculation for concentrations... So if we think back to AS Foundation, we might recall:

First, we use this to work out the number of moles of HCl: (25 x 0.1) / 1000 = 0.0025 mol.

Then we use this with the new volume in the mol equation, in order to calculate the new concentration.

0.0025 mol = ( M x 100cm3) / 1000

2.5 = M x 100

M = 0.025 mol dm-3

Now we have our concentration of [H+] ions, we can put them into the log calculation:

pH = - log100.025

pH = 1.6

There are shortcut methods to finding a new concentration, but if you're unsure of those, just stick to this old method.

Strong bases

Although the log calculation for the pH is exactly the same for strong bases as acids, there's an extra step when you're faced with an aqueous solution of a base.

The water in the solution ionises:

H2O H+ + OH-

At 25oC, the equilibrium lies to the left, and the ions on the right have equal concentrations of 10-7 mol dm-3. As the concentration of H2O is virtually unchanged, we can treat it as a constant. This means we can replace it with a new equilibrium constant, called Kw.

So, just what is Kw?

Kw = [H+(aq)][OH-(aq)]

and at room temp, Kw is 1 x 10-14 mol2 dm-6.

So how does this help us with a base equation? Quite simply, if we put the concentration of the base into the Kw equation, we can work out the concentration of [H+] ions. Like the strong acids, though, we need to assume that the base fully dissociates.

For example, if we have 0.01 mol dm-3 of NaOH, we can use the Kw equation:

Kw = [H+(aq)][OH-(aq)]

We can put in values for Kw and NaOH:

1 x 10-14 = [H+] x 0.01

Rearranging, we get:

[H+] = 1 x 10-14 / 1 x 10-2 = 1 x 10-12

Now, stick this into the log formula:

pH = - log10(1 x 10-12)

pH = 12

And that's pretty much it for strong bases!

Weak acids

As we know, weak acids are only partially dissociated. This means that the concentration won't equal the [H+] concentration and we can't simply use the log equation to find the pH.

Instead, we use something called Ka, the acid dissociation constant. Ka measures how much of the acid dissociates by dividing the concentration of undissociated, by the concentration of ions. If HA(aq) H+(aq) + A-(aq) represents a general equilibrium, Ka will look like this:

The higher Ka, the higher the [H+], so the stronger the acid.

In order to calculate the pH using Ka, we need to make two assumptions:

  • that the [HA] at equilibrium will be pretty much the same as when it started (as hardly any [HA] will dissociate
  • that the concentration of [A-] = [H+]

From this, we can calculate pH. For example, if we have 0.1 mol dm-3 of CH3COOH, and the Ka = 4.8 x 10-5, we use the following method:

Draw up a table of changes from the start to equilibrium. Call the unknown amount of ions x:

  HA H+ + A-
time = 0 0.1 mol dm-3 0   0
time = equilibrium (0.1 - x) x   x
  ˜ 0.1        

Because x is a tiny amount, it doesn't really affect how much HA concentration there is.

Next, we use the Ka equation and fill in what we know:

x2 = 4.8 x 10-6

x = 2.19 x 10-3

Now we've found x, the H+ concentration, we can use it in the log calculation:

pH = - log10 2.19 x 10-3

pH = 2.7

Maybe weak acids take a little more thought, but that's all there is to it!

pKa

Because Ka is really small, it's often easier to compare weak acids by using pKa. This means sticking the Ka into another equation:

pKa = - log10Ka

The values that come out work the opposite of Ka, meaning the smaller the pKa value, the stronger the acid.

If we're given a pKa value, and we need to find the pH of the acid, we can convert pKa back into Ka by using the following equation:

Ka = 10-pka

So, if we're given a Ka of 2.9 x 10-3 for HA, the pKa would be:

pKa = - log10Ka

pKa = - log10 2.9 x 10-3

pKa = 2.54

If we know the pKa = 9.9, the Ka would be:

Ka = 10-pka

Ka = 1.26 x 10-10

Nothing to it!

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Buffers

Learn the proper definition of a buffer, not the one you might learn in biology: buffers are solutions that resist changes in pH. They are very important for anything biological, and we wouldn't be alive if we didn't have them inside of us!

There are two main types of buffers: ones that use weak acids and ones that use weak bases. For both, you need equal concentrations of the acid/base and its salt.

An example with CH3COOH and CH3COONa:

CH3COOH CH3COO- + H+

CH3COONa CH3COO- + Na+

In the two equilibriums, we ignore the Na+ because it's just a spectator ion. We assume that the salt fully dissociates and that the acid doesn't really. So, there will be lots of CH3COOH, from the first equilibrium and lots of CH3COO- from the second equilibrium.

If H+ ions are added (an acid is added), they react with the CH3COO- to form CH3COOH. Because the acid is only weak, it doesn't dissociate much, so the H+ concentration pretty much stays the same and the pH stays much the same...change is resisted.

If OH- ions are added (an alkali), they react with the H+ to form H2O. The weak acid equilibrium then moves to the right hand side to resist change (through Le Chatalier's principal) and restores the lost H+ ions, maintaining pH.

For weak base reactions, let's take the following example:

NH3 + H2O NH4+ + OH-

NH4Cl NH4+ + Cl-

As with Na+, the Cl- is just a spectator ion, so we can ignore it. We assume the weak base doesn't really dissociate, but the salt dissociates completely meaning we have lots of NH3 from the first equilibrium and lots of NH4+ from the second equilibrium.

If H+ is added, this reacts with the OH- to form H2O. The equilibrium is disrupted and thus moves to the RHS through Le Chatalier, to restore the balance. Thus, the OH- concentration is restored, and pH change is resisted.

If OH- is added, it reacts with the NH4+ to form NH3 + H2O. Because NH3 is a weak base, it doesn't dissociate into many ions, so the OH- concentration stays roughly the same and pH change is resisted.

Buffer pH

To work out the pH of a buffer solution, we use the Ka equation; where the [HA] was before, we use the acid concentration, and where [A-] was before, we can replace it with the concentration of the salt.

So:

We can rearrange this:

From this, we just use the old pH = - log10[H+] to work out the pH of the solution.

For example: What is the pH of a buffer solution containing 0.02 mol dm-3 of CH3COOH and 0.05 mol dm-3 of CH3COONa, where the Ka = 1.7 x 10-5 mol dm-3?

We use the buffer equation:

[H+] = 6.8 x 10-6

Next, stick the [H+] into the log equation:

pH = - log10 6.8 x 10-6

pH = 5.17

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Indicators

We've been using indicators of various types for years, to find the pH or the acidity... but just what are they? They can be either weak acids or weak bases, that undergo a colour change when ions or undissociated. In represents the indicator and the equilibrium for a weak acid indicator may be written as:

HIn(aq) H+(aq) + In-(aq)
Colour I   Colour II

The pH range over which an indicator changes colour is equal to pKa=1.

If the pH of the solution is more than one below the the pKa, the colour of the indicator is colour I. If pH is at least one greater than the pKa, then the colour is II.

The colour change occurs, because if we flood with H+, the equilibrium with move to the LHS to decrease the increase, appearing colour I. If we flood with OH-, these will react with the H+ to form water, and the equilibrium will move to the RHS to replenish the loss, appearing colour II.

Two common indicators are phenolphthalein and methyl orange (fortunately, you don't need to know the names!).

Phenolphthalein has a pKa of 9.3. At less than pH=8.3, it will be colourless and at more than pH=10.3, it will be red. In the middle section, it will appear as a pale pink.

Methyl orange has a pKa of 3.7. At less than pH=2.7, it will appear red, and at more than 4.7, it will appear yellow. Between these two figures, it will be orange.

In a titration, indicators can be used to show pH change, if the rapid change in pH coincides with the colour changes of the indicator.

With a strong acid and strong base, the titration curve would look like this:

Because the pH goes from 1 to 14, it doesn't matter which indicator used, as they will all fit into the wide range of the rapid change, as shown to the right.

 

A weak acid and strong base might look like this:

For this, the starting pH is too low for methyl orange's range, as shown to the right. Between the two, only phenolphthalein would change colour with the rapid change of pH.

A strong acid a weak base:

Though both would change colour, the phenolphthalein would only just be in range, so it's better to use methyl orange.

A weak acid and a weak base:

Phenolphthalein is only just in range and methyl orange isn't. It would be better to use another indicator around pKa=7.

In order to work out when the sudden change of pH occurs, we do a simple moles equation. For example: When does the pH change occur if 25 cm3 of 0.01 mol dm3 HCl is titrated with an excess of 0.005 mol dm3 NaOH?

We think about the equation: 1 mol of HCl reacts with 1 mol of NaOH.

Next, we work out how many moles we have of HCl, using the moles in solution equation:

mol = 25 x 0.01 / 1000

mol = 0.00025

So, if we have 0.00025 mol of HCl, we will need 0.00025 of NaOH to neutralise.

The NaOH has concentration of 0.005 mol dm-3, so we use the moles equation again:

0.00025 = 0.005 x ? / 1000

0.25 = 0.005 x ?

? = 50 cm3

The rapid change of pH will occur when 50 cm3 of NaOH is added to the HCl.

The exam may throw any weird and wonderful indicators at you, but you don't need to learn any! All you need to know is how to use them, based on their pKas, and how to draw a titration curve for strong and weak acids and bases.

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That's it for Unifying Concepts... email if you think something's missing. All that's left to say is prepare for lots of moles equations, and good luck!