Advanced Higher Chemistry 2014/2015

answers


could someone please explain?

i know the acid and sodium hydroxide react in 1:1 ratio but the question doesnt actually state the concentartion of each so I assumed the conc of each is equal but i dont know if you can do that

then I said if half of acid is neutralised that is 12.25 cm³ and assuming equal concentraion of acid and alkali that means this will happen when
12.25 cm³ of sodium hydroxde has been added so I went to 12.25 on x axis and read a pH of 4 of the graph which means pK_a = 4

but the answers say
pK_a = 3.75 - 3.8

what am I doing wrong?

thanks I really apprecaite any help
exam on thursday ;(

The vertical part of the graph is where the equivalence point is reached, and thus is the point at which 100% of the alkali has been neutralised.
It follows that the point where half of the acid is neutralised is when half of the volume of alkali needed to neutralise it has been added - simply take half of the value for which the acid is fully neutralised.
This halfway point lies on the horizontal part of the graph, between X and Y ish. This gives you the 3.8 pKa value you're looking for.

could the answer not be B or D here?

the actual answer is B

thnx

This is a bit of a trick question. Answer D could be a hydride, however you know lots of compounds with alkaline pH other than ionic hydrides. The only thing that will ever give you hydrogen gas at the positive electrode is a compound containing negative hydride ions, so to test for hydride ions you'd do B.

So I was flicking through past papers again and I came up against the final question of the 2014 paper.
I don't know why, but I can't seem to get the concentration of the acid right for the buffer solution.

Question a) ii) : http://puu.sh/i1KOt/6fec9dcded.png
Answer: http://puu.sh/i1M51/652742108d.png

I'm fine with absolutely everything else in the answer scheme for this question other than the value for the concentration of the acid.

0.02L of 1mol l-1 KOH gives 0.02mol KOH.
0.04L of 1mol l-1 ethanoic acid gives 0.04mol ethanoic acid.

Acid(monoprotic) + Alkali ==> Salt + Water
We have 0.02mol ethanoic acid reacting with 0.02mol KOH, giving 0.02mol salt, 0.02mol water, and 0.02mol leftover ethanoic acid.

The solution overall is 60cm3 ==> 0.06L, and as C=N/V, both acid and salt should have the same concentration as the volume is the same and their number of moles are the same. This is clearly not the case, as in the answer scheme the concentration of acid is 0.666mol l-1, double that of the salt. I must be going wrong in something I've done above, and I can't find it for the life of me.
The rest of the solution from there on is fine, but can anyone offer help on what I've done above?

ANSWER


im confused with this question

i know an acid donates a proton (H+) ion and the species left is the conjugate base but i can't see how from this equation something has gained only an H+ ion?
thnx for any help

i know it gives the wrong answer but why can't the relationship [H+] = root(K_a c) not be used where we'd use K_a of ethanoic acid from data bok and conc of 0.666666666667 mol/l ?

How do I know if this is nucleophilic or electrophilic. The answer is D.

Don't really know much about unit 3, can someone explain this? The answer is B.

You have a water molecule gaining a proton to become H30+, or the H+(aq) ion. This makes water the base in this reaction, and thus CO2 the acid.
On the right hand side, we see CO2 after losing its proton (look at it as if it has an OH- group on it now - it went from CO2 to HCO3-) as HCO3- and water after gaining its electron as H+ or H3O+.
On the right hand side, one of these can act as an acid and one can act as a base. That which acts as the acid is the conjugate acid and that which acts as the base is the conjugate base.
H3O+ donates a proton to become H2O, and so it is the conjugate acid.
HCO3- can accept a proton to become CO2 (+ H2O) and so is the conjugate base.

Note that HCO3- is amphoteric as it can donate another proton to become the carbonate ion, CO3^2-.

This one confused me a bit at first too, but this is how I puzzled it out.

http://www.new.chemistry-teaching-resources.com/Resources/PastPapers/RevAdvH/AH_Chemistry(Revised)_all_2014.pdf

In question 12 in the mult choice, why is the Answer B ? I thought if X was added that would mean through equilibrium it favours the left hand side?