I would like an explanation of this so I can understand it better for when it sh

Question

I would like an explanation of this so I can understand it better for when it shows up on the exams.

I've been plugging in everything to the equation but I'm still getting it wrong.

I use Ka1 = 2.4x10^-4
Ka2 = 3.8 x 10 ^-11
Kw = 10^-14
[HA] = 0.166M x 2 = 0.332

I keep getting 9.55 x 10^-8

I don't understand why it's wrong :(

Given a diprotic acid, H2A, with two ionization constants of Ka1 2.4x 104 and Ks2 3.8x 10 calculate the pH for a 0.166 M solution of NaHA Number pH9.55x 10-5 Previous 3 Give Up & View Solution Try Again NextExit- Explanation Note that HA is the intermediate form of the diprotic acid. It is an amphiprotic species that can both accept and donate a proton. The following equation is for calculating the H concentration, where you assume [HA1 F, the formal concentration. (When [H2A] and [A2-] are more than 1% of F. [HA] is not equal to F and you should calculate a new value for [HA-1 from [HA-1-F-[HAI-1A2-1 and recalculate rh"1, repeating until a constant value ofHT is reached.)

Explanation / Answer

Answer – We are given, [NaHA] = [HA-] = 0.1666 M,

Ka1 = 2.4*10-4 , Ka2 = 3.8*10-11

We are given the formula for calculating the [H+] when the species acts as amphoteric in nature.

You calculate the [H+] that is 9.55*10-8 M and it is like bellow

[H+] = (Ka1*Ka2*[HA-] + Ka1Kw / Ka1 + [HA-]

        = 2.4*10-4 * 3.8*10-11* 0.166 + 2.4*10-4 * 1*10-14 / 2.4*10-4 +0.166

        = 9.55*10-8 M

We need to calculate the pH and we know formula

pH = -log[H+]

     = - log 9.55*10-8 M

     = 7.02

So answer is 7.02 and not 9.55*10-8 M

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