Question

chem

Answer 1

@Sword

Answer 2

Given any acid (such as HA), we say that it will dissociate in solution by donating, or 'detatching' its H+ (proton) to form it's conjugate base A-.

\[HA _{(aq)} \rightarrow \leftarrow A ^{-} _{(aq)}+ H _{(aq)}^{+} \]

This is an equilibrium system in which a certain amount of HA acid is dissociating (forward reaction) and a certain amount is being reformed once again in the opposite direction (reverse reaction). Equilibrium is when the rate of the forward and reverse reactions are the same.

Ka is the acid dissociation constant, a measure of the strength of acid (i.e. it's ability of dissociate/be a good proton donor over its ability to reform). In a similar manner to the equilibrium constant (Keq), we calculate Ka taking the ratio of the concentrations of the products and reactants art equilibrium.

\[K _{a} = \frac{ [A ^{-}][H ^{+}] }{ [HA] } ([...]~means~concentration)\]

So, the stronger our acid, the more readily it will dissociate (and the more stable its conjugate base) and thus the greater our concentration of products at equilibrium, meaning a larger Ka value.

pKa is a scale (a bit like pH) which we can also use to examine the strength of an acid using more manageable numbers. So, given a Ka value for an acid's dissociation, its pKa value will be:

\[pK_{a}= -\log _{10}K _{a}\]

This logarithmic scale means that the larger our Ka value, the smaller our pKa value, and vice versa. So to identify the strongest acid, we need to find the one with the lowest pKa value, corresponding to the highest Ka value and a greater readiness to form its conjugate base and dissociate.

Answer 3

Okay, so what do you think the answer is?

Answer 4

Well, if you read through my last post hopefully I've explained how you can figure out the answer yourself, so which one do you think it is @vera_ewing ?

Answer 5

I think A. HCl, but I'm not really sure.

Answer 6

What's your reasoning behind thinking it's HCl @vera_ewing ? Can you tell me why HCl might form the most conjugate base at equilibrium?

Answer 7

Well doesn't it have the lowest pKa value?

Answer 8

Yes, so that's why it must have the highest Ka value and most amount of H+ and conjugate base (Cl-) formed at equilibrium. Hope that helped you out and you're able to understand it a little bit better now!

Answer 9

So the answer is A?

Answer 10

Based on this reasoning, it should be A....but please take the time to read through the replies @vera_ewing and then use the information to try and decipher the answer the answer for yourself in future!