A2 level. ionic equilibria help.

[cashew]

HI, I've been having problems in constructing ph curves lately.
how do i solve this question?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_4.pdf
Q1, part b(iii)
here is the mark scheme
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_ms_4.pdf
I am not getting the ph around 5 when 5cm3 of the base is added. im getting the ph around 2.8. which is not even close to 5.
I'll appreciate any help.
thanks in advance.

 

[nehaoscar]

HI, I've been having problems in constructing ph curves lately.
how do i solve this question?
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_4.pdf
Q1, part b(iii)
here is the mark scheme
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_ms_4.pdf
I am not getting the ph around 5 when 5cm3 of the base is added. im getting the ph around 2.8. which is not even close to 5.
I'll appreciate any help.
thanks in advance.

Firstly from part ii you know the pH before the base is added is 1.94 so the graph starts from 1.94

Then the acid and base react in a 1:1 ratio so from the question we know:
10 cm3 of 0.10 mol dm−3 acid = 10*0.1 = 1mol of acid
so using v = n/c for the base => 1/0.1 = 10cm^-3 (since c of base is given as 0.1)
so now we know that the end point is at 10cm^-3

Now at the end of the titration we find the pH of the base by using
pH = -log[H+]
this gives 13
so the graph ends at point 13 pH at 20cm^-3

Now since this is a weak acid - strong base graph, the end point line (i.e the straight part of the line) should be between 7.5 - 11.5

So to summarize plot these points:
1. pH 1.94 at 0cm^-3
2. pH 13 at 20cm^-3
3. end point straight line from pH 7.5-11.5 at 10cm^-3
4. Join all these points to obtain a sketch of the graph

As you have asked at the point 5cm^-3 you're not getting pH 5 remember that it's a rough graph and you can't calculate the pH at every value
You just need to know the end points for the different acid-base titrations (i.e strong-strong , strong- weak etc...) which are given in the text book so that when you sketch it will be roughly the same as shown in ms
Hope this helps

 

[cashew]

Firstly from part ii you know the pH before the base is added is 1.94 so the graph starts from 1.94

Then the acid and base react in a 1:1 ratio so from the question we know:
10 cm3 of 0.10 mol dm−3 acid = 10*0.1 = 1mol of acid
so using v = n/c for the base => 1/0.1 = 10cm^-3 (since c of base is given as 0.1)
so now we know that the end point is at 10cm^-3

Now at the end of the titration we find the pH of the base by using
pH = -log[H+]
this gives 13
so the graph ends at point 13 pH at 20cm^-3

Now since this is a weak acid - strong base graph, the end point line (i.e the straight part of the line) should be between 7.5 - 11.5

So to summarize plot these points:
1. pH 1.94 at 0cm^-3
2. pH 13 at 20cm^-3
3. end point straight line from pH 7.5-11.5 at 10cm^-3
4. Join all these points to obtain a sketch of the graph

As you have asked at the point 5cm^-3 you're not getting pH 5 remember that it's a rough graph and you can't calculate the pH at every value
You just need to know the end points for the different acid-base titrations (i.e strong-strong , strong- weak etc...) which are given in the text book so that when you sketch it will be roughly the same as shown in ms
Hope this helps

Thankyou very very much!