pH............. help me please with this exam question?

Question

a) what do you understand by the terms strong acid and weak acid and how is the species H3O+ classified?
b) what is meant by the pH of solution, stating how this quantity is defined and the approximation that is usually made in applying it to a solution. how may pH be calculated for a solution of concentration c of
i) a monoprotic stong acid and
ii) a monoprotic weak acid with an acidity constant Ka?
hence calculate the pH of 0.05 mol dm-3 HCL solution and 0.05 mol dm-3 ethanoic acid solution (Ka = 1.8 * 10-5 mol dm-3)
c)describe with a diagram how the acidity constant of a weak acid may be determined by a pH titration and explain the theoretical basic for this determination
d) give two reasons why it is difficult to measure the acidity constant for polyprotic acids such as polypeptides.

Answer 1

strong acid is one that fully dissociates in aqueous solution, weak acid forms an equilibrium where it partially dissociates.

pH = -log [H+] where [H+] is the molar concentration of H+.
it is used for simplifying the quantity for easier use.

monoprotic strong acid takes one step to dissociate fully
monoprotic weak acid takes one step to partially dissociate in equilibrium

Ka = [H+][A-]/[HA]

polyprotic acids take more than one step to dissociate
e.g. bicarbonate

I'm not sure what d. is, but for c, a titration only measures amount of base it takes to neutralize an acid. By doing this you could calculate the concentration of the acid and from that calculate the pH using the formula M1V1=M2V2.

Answer 2

The other answerers have answered parts (a) and (d), so I'll just concentrate on the calculation part.

For 0·05M-hydrochloric acid, you assume it to be fully ionised. Therefore the value of [H+] will be 0·05. Hence, the pH will be -log0·05 , which equals 1·301 or 1·30 , to 2 decimal places.

For ethanoic acid, you start by writing out the basic equation: [H+][A-]÷[HA] = Ka
For 0·05M-ethanoic acid, since it is a weak acid, you can assume, to a first approximation, that the concentration of unionised acid [HA] = 0·05
You can also assume that [H+] = [A-] (neglecting the contribution to [H+] from the water itself)
Thus: [H+]² ÷ 0·05 = 1·8 x 10^(-5) ;
therefore [H+]² = 0·05 x 1·8 x 10^(-5) = 9 x 10^(-7) ;
therefore [H+] = √(9 x 10^(-7))
and the pH will be -log(√(9 x 10^(-7))) .
I haven't got my calculator with me, but I think this comes to 3·023 or, to 2 decimal places, 3·02 .

Were you really expecting answerers to draw the diagram for part (c) ? Unfortunately, one can't draw diagrams on this format !

Answer 3

a) Strong acids are 100% ionized into H+ and A-. Weak acids are less than 100% ionized. H3O+ is the conjugate acid of water and is the strongest acid that can exist in a water solution. If there were astronger acid in water, it could react with water to produce H3O+.

b) pH = -log [H+] The approximation is that pH is really -log a(H), where a(H) is the activity of hydrogen ion. The activity is the concentration multiplied by the activity coefficient for the particular situation.

To calculate pH for a strong monoprotic acid, HX, take -log[HX]. For a weak acid, pH = -log[H+] = -log{Ka[HX]}^1/2. That is -log of the square root of Ka[HX].

c. If you titrate such that [HX] = [NaX], then the pH at the end point is the pKa = -logKa.

d. There are many Ka's for polypeptides, because they come from all the different acidic amino acids: glutamic acid and aspartic acid. Also, once you take the first H+ off one of these amino acids, it changes what the Ka of the second one would have been.

Answer 4

Yea, the different individual is real. better of our training are in preserving with round table discussions, and for that reason- were better in preserving with very large paper (this very last exam era, I probably wrote about 40 pages for my 3 training. There are favourite distinct determination checks in some training, yet no longer very many.