Chemical amounts (A2)

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I've been given five questions on titration in class and find it immensely difficult to solve. Would you please be able to help me.
 
The question goes like this: 20 cm³ of 0.1 mol dm^-3 of ethanoic acid is titrated with 0.1 mol dm^-3 KOH. K_a= 1.74 x 10^-5.
 
1) pH at the start before any KOH added
2) pH when 10 cm³3 of KOH added
3) pH when 18 cmcm³ of KOH added
4) pH when 20 cmcm³ of KOH added
5) pH when 30 cmcm³ of KOH added
 
Could you please explain why the answers are so as well because my teacher tried to explain it but I still could not get it. 260308

Jenkin writes ....
 
(1) is to calculate the pH of 0.1 mol dm^-3 ethanoic acid. K_a for this weak acid is 1.74 x 10^-5.
 
In aqueous solution, ethanoic acid is slightly dissociated (ionised). I shall use HA to mean ethanoic acid: the equilibrium is
HA <=> H⁺ + A^-
 
so K_a = ([H⁺][A^-])/[HA] = 1.74 x 10^-5
 
[H⁺] and [A^-] are equal (ignoring the very small number of H⁺ ions arising from dissociation of water molecules). [HA] = 0.1 mol dm^-3 (ignoring the small number of HA molecules which have ionised; the point is that it is very small compared to 0.1)
 
So now ([H⁺]²)/0.1 = 1.74 x 10^-5, and therefore [H⁺]² = 1.74 x 10^-6.
 
Taking square roots gives [H⁺] = 1.319 x 10^-3), from which pH = -log (1.319 x 10^-3 = 2.88
 
(2) On adding KOH, the reaction is CH₃COOH + KOH -> CH₃COOK + H₂O. I shall write this as HA + KOH -> KA + H₂O.
Since both acid and alkali are of the same concentration, the 10 cm³ of alkali will neutralise 10 cm³ of the acid. So we will now have half the original number of moles of the acid. Originally, there were (20/1000) x 0.1 moles of HA (i.e. 2 x 10^-3). Now there are 1 x 10^-3 as half have been neutralised.
 
But the total volume of the liquid has increased from 20 to 30 cm³ and so the [HA] is now
= (1 x 10^-3 x 1000)/30, which is = 0.0333 mol dm^-3.
 
The other thing to notice now is that the KA (potassium ethanoate) formed in the reaction is completely ionised. There will be 1 x 10^-3 moles of it in the 30 cm³ of solution, so the [A^-] is also going to be 0.0333 mol dm^-3.
 
So K_a = ([H⁺] x 0.03333)/0.0333 = 1.74 x 10^-5, and therefore [H⁺] = 1.74 x 10^-5, and so pH = 4.76.
 
(3) After the addition of 18 cm³ of the KOH, we can calculate the pH in the same way as in (2). Now 18 cm³ of the acid will have been neutralised, so there will be 1/10 of the original number of moles (0.2 x 10^-3) of acid left, and the total volume of solution is now 38 cm³.
 
So the [HA] is now (0.2 x 10^-3 x 20)/38 = 1.0526 x 10^-4 mol dm^-3.
 
The amount of A^- will be 1.8 x 10^-3 mol, and thus [A^-] is (1.8 x 10^-3 x 1000)/38. If we now substitute these values in the expression for K_a and then find the [H⁺], this gives the pH of 7.41
 
(4) When 20 cm³ of KOH have been added there is no more HA present. This makes things more difficult: in effect, we now have to calculate the pH of a 0.05 mol dm^-3 solution of KA. There is no very simple way of doing this; it involves using K_w, the ionic product of water, K_w = [H⁺] x [OH^-] = 1 x 10^-14, and also the idea of salt hydrolysis. You could look this up in a textbook, but what it amounts to is that:
 
[H⁺] = K_a x (Kw/Ka)^0.5, which, for this example, gives a pH of 9.38
 
(5) After adding 30 cm³ of KOH, we have now got a solution of KOH of concentration
(0.1 x 20)/30 mol dm^-3.
 
KOH is a strong base, so it is completely ionised and the [OH^-] is the same as [KOH].
To find the [H⁺] we need to use K_w again:
 
K_w = 1 x 10^-14 = [H⁺] x [OH^-], which gives a pH of 12.82
 
Hope this helps. If not, come back to us via Re:act, not email.

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updated: 28 March 2008