Acid-Base Equilibria (3) Defining and using Ka and Kw

Edexcel A level Chemistry (2017)

Topic 12: Acid–Base Equilibria:

Here are the next five learning objectives:

Defining and using K_a and K_w

12/9. To be able to deduce the expression for the acid dissociation constant, Ka, for a weak acid and carry out relevant calculations

12/10. To be able to calculate the pH of a weak acid making relevant assumptions

12/11. To be able to define the ionic product of water, Kw

12/12. To be able to calculate the pH of a strong base from its concentration, using Kw

12/13. To be able to define the terms ‘pKa’ and ‘pKw’

12/14. To be able to analyse data from the following experiments:

i  measuring the pH of a variety of substances, e.g. equimolar solutions of strong and weak acids, strong and weak bases, and salts 


ii  comparing the pH of a strong acid and a weak acid after dilution 10, 100 and 1000 times

12/15. To be able to calculate Ka for a weak acid from experimental data given the pH of a solution containing a known mass of acid

If pH is an unreliable measure of acid strength then chemists turn to K_a and use that value.

K_a is called the acid dissociation constant.

If a weak acid dissociates like this:

CH₃COOH    +H₂O    ⇌   CH₃COO—   +    H₃O+

Then Kc can be written:

but it is pretty safe to assume that the concentration of water remains constant during the dissociation of the weak acid since it is hardly dissociated and the position of equilibrium is well over to the left. 

Therefore this relationship leads to a second constant which is called the acid dissociation constant K_a.

The acid dissociation constant gives us a direct measure of the extent of dissociation of the hydrogen ions and therefore of the strength of the weak acid. 

There are tables of K_a values on the internet.

Let's see how to use K_a to calculate the pH of a solution of a weak acid.

What is the pH of a 0.1M solution of chloroethanoic acid (CH₂Cl.COOH)?

CH₂Cl.COOH   +    H₂O   ⇌   CH₂Cl.COO^—   +   H₃O⁺

But we can also say that [CH₂Cl.COO^—]  must equal [H₃O⁺ ]

From the equilibrium equation.

So we can re write the Ka equation like this:

What’s more since this is the dissociation of a weak acid it is reasonable to assume that the concentration of the acid hardly changes on dissociation.

So [CH₂Cl.COOH] the undissociated acid is the same concentration at equilibrium which is 0.1M.

Therefore:

rearranging gives:

[H₃O+]    =     √1.4× 10^–4

[H₃O+]   =   1.18  × 10^–2  mol.dm^—3

and from this

pH   =    — log₁₀[H₃O⁺]

pH   =   1.93

Now the best way to get your head round this is to do as many calculations as you can.

Here are a few to get you started:

What is the pH of

0.1M ethanoic acid (CH_3.COOH)

0.1M hydrogen cyanide (HCN)

0.1M chloroethanoic acid

0.45M benzoic acid (C₆H_5.COOH)

You can find the relevant K_a values here on the internet..

So what happens if we apply a similar argument to water itself?

Water ionizes itself like this

H₂O    +     H₂O     ⇌     H₃O+    +     OH^—

And for this equilibrium K_c be:

Now in either acidic or alkaline conditions the concentration of water remains virtually constant being so high compared with the concentrations of the other species (it is over 50M). 

So incorporating the water term in the constant gives us K_w the ionic product of water. 

K_w  =   [H₃O⁺]×[OH^—]

And if we take negative logs to base 10 of this expression we are left with a very convenient term.

pK_w   =    pH   +   pOH  =  14

for most aqueous solutions of weak acids and bases at 298K.

From this we can also see that at 298K

[H₃O⁺]   =   10^—7  mol.dm^—3

and  therefore for water at 298K

pH   =   7

So let’s calculate the pH of a solution of a strong base: 0.1M sodium hydroxide solution.

This solution will have a hydroxyl ion concentration of 0.1M.

Now

pH  +   pOH   =14

 pOH  =   —log ₁₀ [OH^—]

pOH   =   —log ₁₀(0.1)  =  1

pH   =   14  —   1   =   13

Finally, in this long post let’s describe how to calculate K_a of a weak acid given its solution concentration in grams per litre.

Suppose that the pH of a solution of ethanoic acid is 3.4O and we have 250ml that contain 0.15g of the acid. 

The concentration of the acid is 0.6 g.dm^—3

And that is 0.6/60 =  0.01 mol.dm^—3

From the pH was can find the [H₃O⁺] since [H₃O⁺]  =    10^—pH

[H₃O⁺]   =   10^—3.4    =  0.000398 mol.dm^—3

from this we can construct the K_a value since we know [H₃O⁺] and can assume that [CH₃.COOH] =  0.01M  as it is a weak acid and is therefore hardly dissociated.

Therefore      

1.58   ×  10^—5 mol.dm^—3