The Oceans F335.

Acids vary in have strength. Strong acids have a higher ability to donate protons than weak acids. Strong acids also fully dissociate in water whereas weak acids only partially dissociate in water as they have a lower tendency to donate protons. Examples of strong acids are hydrochloric, sulfuric and nitric acid. Examples of weak acids are carboxylic acids and carbonic acid. 

Ka is the acidity constant. The greater the value of Ka the stronger the acid. Ka can be converted into a value known as pKa which simplifies the value of Ka. The smaller the pKa value the greater the acid strength for example:

Ka methanoic acid = 0.00016 

pKa methanoic acid = 3.8 

pKa = -logKa 

Ka = 10^-pKa

pH can be caluclated using the following equation: pH = -log[H+(aq)]

Strong Acids: As strong acids fully dissociate, we can assume that the concentration of acid put into the solution is the same as the concentration of hydrogen ions. Calculate the pH of a 0.001mol dm^-3 solution of hydrochloric acid:

[H+] = 0.001 so pH = -log 0.001 = 3.00 

Weak Acids: The concentration of the acid and the Ka are needed to calculate the pH of weak acids as only some of the acid as dissociated and so the concentration of hydrogen ions and of the acid put into the solution will be different. Calculate the pH of 0.01mol dm^-3 ethanoic acid (Ka = 0.000017) 

0.000017 x 0.01 = 1.7 x 10^-7 

√1.7 x 10^-7 = 4.1 x 10^-4 = [H+] 

Thus pH = -log(4.1 x 10^-4) = 3.4

Strong Bases: The concentration of the base and the concentration of [OH(aq)] is the same in strong bases as the base has fully dissociated in aqueous solution. To calculate the pH of strong bases, the following is required: 

Kw - The ionic product of water constant. The value for this is always 1 x 10^-14 mol^2 dm^-6

Calculate the pH of a 0.01mol dm^-3 solution of NaOH: 

The strong base has fully dissociated therefore [OH(aq)] = 0.01mol dm^-3

Kw = [H+(aq)] [OH(aq)] 

[H+(aq)] = 1 x 10^-14/0.01 = 1 x 10^-12

pH = -log (1 x 10^-12) = 12.0 

Buffers are solutions which are able to resist changes in pH on the addition of small amounts of acid or alkali. They're made up of:

• A weak base and one of its salts OR
• A weak acid and one of its salts 

This equations shows ethanoic acid acting as a buffer. There is plenty of the acid available which hasn't dissociated which can dissociate if required when alkali is added to the solution to provide more H+ ions and reverse this effect. Likewise, there is plenty of A- (CH3COO-) available to combine with the H+ ions to form more HA (CH3COOH) of there is acid added to the solution. The concentration of H+ stays roughly constant. 

It's important that buffer solutions contain both proton donors (acid) and proton accceptors (the ion of the acid e.g CH3COO-).

All calculations with buffers require use of the following equation: 

Ka = [H+(aq)] x [salt]/[acid]  

If the Ka of the acid is known as well as the concentrations of both the salt and the acid then the pH can be calculated:

Calculate the pH of a buffer solution made with equal volumes of 0.20mol dm^-3 ethanoic acid and 0.10mol dm^-3 sodium ethanoate solutions (Ka ethanoic acid = 1.7 x 10^-5)

By mixing equal quantities each original concentration will be halved because when solutions are mixed their individual concentrations are reduced in proportion to their volumes: 

[H+(aq)] = (1.7 x 10^-5 x 0.10)/0.05) = 3.4 x 10^-5 mol dm^-3

pH  = -log (3.4 x 10^-5) = 4.5

Weak Acids

Assumption 1: [H+(aq)] = [A-(aq)]: H+ can also be formed through the dissociation of water, however in weak acids this produces far fewer H+ ions so the hydrogen ions produced by the dissociation of water is neglible. 

Assumption 2: HA at equilibrium = Initial HA: The amount of H+ lost when HA dosociates can be neglected because weak acids don't fully dissociate and this fraction is very small. 

Buffer Solutions

Assumption 1: All the A- ions come from the salt: The weak acid supplies very few A- ions as it doesn't fully dissociate. 

Assumption 2: Almost all HA molecules remain unchanged: The high concentration of A- in the solution pushes the equilibrium further to the left, meaning the concentration remains virtually unchanged. (Similar to assumption 2 for weak acids)