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Acid Base Chemistry

4. Acids
a. Defining strong acids and weak acids
b. Strong acids
i. What makes a strong acid?
ii. How to calculate the pH of a strong acid
c. Weak acids
i. What makes a weak acid?
ii. How to calculate the pH of a weak acid

4cii. How to calculate the pH of a weak acid
[slide 1]
Now we get on to something slightly more mathematical which is calculating the pH of weak acids. So, we are learning to: derive this formula, which is H+ concentration
equals square root of the equilibrium constant, Ka, multiplied by the weak acid concentration. Then we will use pH equals minus log of H+ concentration to calculate the pH of these weak acids: 0.1 mol dm-3 HCN, with a Ka value of 4.9x10-10, and 0.05 moldm-3 phenol with Ka equals 1.28 x 10-10.

[slide 2]
First of all, let's derive this formula, the concentration of H+ equals the square root of Ka multiplied by the weak acid concentration. Ka is the ratio of the products to the reactants in the equilibrium reaction in which HA dissociates into H+ and A-. Here I've shown the H3O+ ion which is formed by reaction of a water molecule with HA, which is strictly accurate. However, in the expression for Ka here I've used the shorthand H+ instead of H3O+, multiplied by the other product A-, and then divided by the concentration of HA, which of course is a reactant. The water is left out of this expression - in fact it is left out of all of these acid base equilibrium constants - because water is so much in excess that it's concentration changes insignificantly as a result of these reactions.

[slide 3]
So in a solution of a weak acid, most of the H-A molecules remain bonded, and a small proportion break apart into H+ and A- ions.
[slide 4]
We can write an equilibrium reaction for this.
[slide 5]
For which there will be a given value of Ka.
[slide 6]
And we can write the expression for Ka as the concentration of the products divided by the concentration of the reactants; that is, H+ multiplied by A- divided by HA.
[slide 7]
However, if we look back at the species in the solution, and count the relative numbers of A- and H+, we can see that they are the same. For every H+ that is created, there will always be an A- created as well, as the HA molecule breaks apart. Therefore, we can say that the concentration of H+ equals the concentration of A-. Now, if we consider the expression for Ka, the H+ and A- concentrations both have the same value. So we can substitute H+ in the place of A-.
[slide 8]
This gives Ka equals H+ times H+, or H+ squared, divided by the HA concentration.
[slide 9]
multiply both sides of the equation by HA gives Ka times HA equals H+ squared. And then we want H+ concentration on its own, so take square roots of both sides...
[slide 10]
....gives our final answer of H+ concentration equals the square root of Ka multiplied by HA. This is what we will need to use when calculating the pH of a weak acid. There is one approximation in using this equation which you should be aware of, and that concerns the value of HA.
[slide 11]
If we use the original HA concentration in this equation, this is a good approximation as long as the concentration has dropped only very slightly. This is the case for small values for Ka, but the approximation is less good as Ka gets larger. Generally Ka is small, though, and this approximation is valid.
[slide 12]
Let's now go on to calculate the pH of some weak acids. First of all, a 0.1 moldm-3 solution of HCN with Ka = 4.9 x 10-10, so a very small Ka value. We know that H+ concentration equals the square root of Ka multiplied by the HA concentration, so that is the square root of 4.9x10-10 multiplied by 0.1. This gives an H+ concentration of 7 x 10-6 moldm-3. Note, that this value is pretty small compared to a strong acid, which would have an H+ concentration of 0.1 moldm-3. In fact it is a lot closer to the H+ concentration of water, which is 10-7 moldm-3.

To work out the pH from the H+ concentration, we do negative log of the H+ concentration, and if you do negative log of 7 x 10-6, that gives a value of 5.15.
[slide 13]
Now, we change the numbers, change the compound, but the method is exactly the same. For 0.05moldm-3 phenol, with Ka = 1.28 x 10-10, we have H+ concentration equals the square root of 1.28 x 10 -10 multplied by 0.05 which gives 2.53x10-6 moldm-3. Then the pH is minus log of this, which gives a pH of 5.6, so once again, slightly acidic.

[slide 14]
OK, a couple of little tests here, the first one on the derivation of the expression for H+ concentration for a weak acid.
[slide 15]
[slide 16]
[slide 17]