Aha i wrote both. Well i just wrote the equation on the side but the words "Potassium acetate" was on the lines so i should be alright.This^^^
Whether it's valid for the question or not is another thing =p
The question said that both acids were monoprotic...By the shape of the titration curve for acid 2 could you assume it was diprotic? It started steep and then tapered off, not smooth like acid 1 suggesting two protons donated.
This would make it a strong, concentrated diprotic acid such as sulfuric?
its not about the steepness of the titration curve, its by finding the equivalence point.Yeah I put the salt as CH3COOK, The first acid was stronger as it had a steeper titration curve, the second was weaker as it had a shallower titration curve. The first was more dilute because it took less to neutralise than the second because it was further left.
yes thats the answer pretty much.oh btw guys, the question said explain the difference in strength and concentration, i.e. identify that acid 1 was stonger than 2 but less concentrated than 2 and hence explain what that meant in terms of acid 1 having more molecules ionised in soln (strength), but acid 2 having more acid molecules per volume of water (concentration)...well thats what my chemistry teacher said and hes a senior hsc marker...
Acid 1 was a less dilute, strong acid. Acid 2 was a more concentrated, weak acid. The volume of KOH showed which was more dilute/concentrated, the pH showed whether an acid was strong or weak. As KOH is a strong base, only a strong acid would have neutralised at pH 7.This is how i done that question.
First you look at the intial pH of both acids, i.e BEFORE the titration.
1 had a ph of 2, and the other of about ~1.2 or smthing.
then you use the formula [H+] = 10^-pH
To find the [H+] concentration. Obviously the acid which had ~1.2 pH was more concentrated.
however that acid was weaker than the one with the pH value of 2. you could tell this by just looking at the graph and comparing the times its took to reach equivalence point.
And hence that's why i wrote KCL as my salt.
thats, what i ment woops, wrote it the other way roundAcid 1 was a less dilute, strong acid. Acid 2 was a more concentrated, weak acid. The volume of KOH showed which was more dilute/concentrated, the pH showed whether an acid was strong or weak. As KOH is a strong base, only a strong acid would have neutralised at pH 7.
Unfortunately, your salt is wrong too. KCl would form from potassium hydroxide and hydrochloric acid. You were asked to name a salt from an acid similar to acid 2. Acid 2 was a weak acid.