Question

Question #0de88

Answer 1

Your weakest acid will be hypochlorous acid, `HClO`.

When dealing with weak acids, all you really need to do in order to have an idea about their relative strength is look at the magnitude of the acid dissociation constant, `K_a`.

The acid dissociation constant tells you how many molecules of a particular acid will ionize in aqueous solution. A very small `K_a` means that the vast majority of the acid molecules will not ionize, i.e. they will not lose their acidic proton.

For a generic weka acid dissociation, you have

`HA_((aq)) + H_2O_((l)) rightleftharpoons A_((aq))^(-) + H_3O_((aq))^(+)`

By definition, the acid dissociation constant is

`K_a = ([A^(-)] * [H_3O^(+)])/([HA])`

If this value is very small, then that means that the concentration of the weak acid, `HA`, is bigger than the product of the concentration of the conjugate base, `A^(-)`, and hydronium ions, `H_3O^(+)`.

So, in order to determine the weakest acid given, look for the smallest `K_a` value. In your case, the smallest `K_a` belongs to `HClO`.

`K_a = 3.8 * 10^(-8)`

The pKa is simply the negative logarithm of the acid dissociation constant.

`pKa = -log(K_a)`

In the case of hypochlorous acid, you get

`pKa = -log(3.8 * 10^(-8)) = color(green)(7.42)`

The weakest conjugate base will belong to the strongest of the acids. The strongest of these weak acids will have the greatest degree of ionization, which implies that it will have the biggest `K_a`.

In your case, the biggest `K_a` value belongs to nitrous acid, `HNO_2`. Its conjugate base is the nitrite anion, `NO_2^(-)`.

`underbrace(HNO_(2(aq)))_(color(blue)("acid")) + H_2O_((l)) rightleftharpoons underbrace(NO_(2(aq))^(-))_(color(green)("conjugate base")) + H_3O_((aq))^(+)`

In order to determine the conjugate base's `pKb`, you first need to calculate its base dissociation constant, `K_b`. The relationship between `K_a` and `K_b` for a weak acid/conjugate base pair looks like this

`K_a * K_b = K_W`, where

`K_W` - the water ionization constant, equal to `10^(-14)`

In your case, the `K_b` of the nitrite ion will be

`K_b = K_W/K_a = 10^(-14)/(4.5 * 10^(-4)) = 2.2 * 10^(-11)`

Thus, the `pKb`, which is the negative log of `K_b`, will be

`pK_b = -log(K_b) = -log(2.2 * 10^(-11)) = color(green)(10.66)`

Answer 2

`K_a`, the acid dissocation constant, is equal to `[H^+][A^-]`/`[HA]`, where `HA` is the undissociated acid and `[H^+]`, `[A^-]` are the conc. of hydrogen ion and the conjugate base `A^-`. It follows that the strongest acid is the one with the highest value of `K_a`, and the weakest acid is the one with lowest such value.

a. The weakest acid is therefore quite clearly `HClO`, hypochlorous acid, the acid with the lowest acid dissociation constant, and its p`K_a` = `-` `log_10` `K_a`. I haven't a calculator to hand and I am sure I can't find the function on this wretched computer.

b. The weakest conjugate base is clearly the counterion of the strongest acid. Because `HNO_2` is the strongest acid, this means that its counjugate base, `NO_2^-`, competes poorly for the proton (and is by definition weakly basic). Now, given that p`K_a` + p`K_b` = 14, all I have to do is calculate the sum 14 `-` `log_10` `K_a` to give p`K_b`.

If you need clarification or if I've made an error let me know.