1. Acids and Bases Calculations

2. Calculating pH of Strong Acids
pH is a measure of hydrogen ion concentration.
A strong acid When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
In the general case . . .
These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen ions that we can think of the reaction as being one-way. The acid is virtually 100% ionised.
For example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the reverse reaction happens that we can write:

3. Strong Acids
At any one time, virtually 100% of the hydrogen chloride will have reacted to produce hydroxonium ions and chloride ions. Hydrogen chloride is described as a strong acid.
To calculate pH use pH = -log10 [H+] where H+ is the concentration of the acid.
Special circumstances: sulphuric acid

4. Sulphuric Acid
Sulphuric acid like many other acids are known as polyprotic.
This is because they have more than one proton to donate.
For sulphuric acid remember that the concentration of H+ ions is twice the concentration of the acid in the question, (i.e. multiply the concentration of the acid by two before logging it).
H2SO4 has two protons and therefore when calculating pH we multiply molarity by 2.
Example:
What is the pH of mol dm-3 sulphuric acid?
Answer: 2 x = mol dm-3
pH = - log 10 [0.0100] = 2.00

5. Calculating the Concentration of Sulphuric acid from pH
If working backwards from pH to find concentration use 10-pH then you must divide the answer you get by two to get the concentration of the acid. (The concentration of hydrogen ions you have worked out is twice the concentration of the acid.)
This would be a 2 mark or 3 mark for sulphuric acid since there is an extra step.

6. Calculating pH of weak acids
A weak acid is one which doesn't ionise fully when it is dissolved in water.
Ethanoic acid is a typical weak acid. It reacts with water to produce hydroxonium ions and ethanoate ions, but the back reaction is more successful than the forward one. The ions react very easily to reform the acid and the water.
At any one time, only about 1% of the ethanoic acid molecules have converted into ions. The rest remain as simple ethanoic acid molecules.
Most organic acids are weak. Hydrogen fluoride (dissolving in water to produce hydrofluoric acid) is a weak inorganic acid that you may come across elsewhere.

7. Acid Dissociation Constant, Ka
In order to calculate pH of weak acids we must know its acid dissociation constant.
This is because of the low % of dissociation.

8. Write out the equilibrium expression (Ka)
convert hydrogen concentration [H+] to pH
Example: What is the pH of 0.1moldm-3 ethanoic acid, which has a Ka of….1.74X10-5moldm-3?)
Equilibrium equation.
CH3COOH (aq) ↔ H+ (aq) + CH3COO-(aq)
So the general Ka expression will be……
Ka = [CH3COO-(aq)][H+(aq)]
[CH3COOH(aq)]
But we need to consider when full equilibrium has been reached and take account of our initial concentration of acid.

9. CH3COOH ↔ H+ + CH3COO- CH3COOH
Start Conc.
0.100
At equilibrium.
[CH3COO-]
[H+]
[CH3COO-]
CH3COOH
For every ethanoic acid molecule that dissociates an ethanoate ion and a H+ also forms. H+
CH3COO-

10. Writing the new Ka expression at equilibrium.
Initial concentration minus the dissociation concentration of the ions.
Ka = [CH3COO-(aq)]eq[H+ (aq)]eq
0.100 – [CH3COO-(aq)]eq
As [CH3COO-] = [H+] at equilibrium we can write…
Ka = [H+]2
0.100 – [H+]
Because the disassociation of the weak acid is so small we assume – [H+] = 0.100

11. Plugging the numbers in..
Substituting in the values gives….
1.74 X 10-5 = [H+]2
0.100
[H+]2= X 1.74 X 10-5
[H+] = √(1.74 X 10-6)
[H+]= X 10-3moldm-3
Go through more step by step examples on the board.
Use worked examples on board from textbook pages 144/145
= pH = (using –log) 11

12. Calculating Concentration of a Weak Acid
[H+] = √ (Ka x [HA])

13. Finding Ka or pKa from pH of a weak acid
The pH of a weak acid, HA, of concentration 0.100mold m-3 was found to be 4.00.
Calculate the pKa for the acid.
pH = 4.00
[H+] = 1.00 x10-4 mol dm-3

14. HA H+ A-
Start concentration
mol dm-3
0.100
Equilibrium concentration mol dm-3
x10-4 =
1.00 x10-4

15. Ka = [H+][A-]
[HA]
= (1.00 x10-4)2
0.0999
= x 10-7 mol dm-3
pKa = -log10 Ka
= 7.00

16. Now try these:
1.) Ethanoic acid mol dm-3 had a pH calculate pKa of acid.
Find Ka and pKa of the following:
2.) mol dm-3 HX it ifs pH is 4.50
3.) mol dm-3 HY if its pH is 3.25
4.) mol dm-3 HZ if its pH is 5.70

17. Answer to no.1 [H+] = 10-pH = 10-4.56 = A Ka = [H+][CH3COO-] [CH3COOH]=
= A
Ka = [H+][CH3COO-]
[CH3COOH]
= [A]2
[B]
= Ka
= pKa = -log10 Ka
CH3COOH H+
CH3COO-
Start concentration
mol dm-3
0.500
Equilibrium concentration mol dm-3
A = B A

18. Quick Quiz! 1.) Calculate pH of 0.056 mol dm-3 H2SO4.
2.) Calculate concentration of HCl pH 3.52.
3.) Write the Ka expression for ethanoic acid in water.
4.) Is ethanoic acid a weak or strong acid?
5.) What is a weak acid?
6.) What is the Ka if pKa value of HNO3 is 4.5 x10-5?
7.) Calculate pKa from Ka value of HCl, 3.2 x10-3 mol dm-3.
8.) What is the pH of 0.56moldm-3 nitrous acid (HNO2), which has a Ka of 1.56X10-3moldm-3?)