[Vicktor]

This problem has been haunting me for 2 days, and I have no clue why I can not solve it:

A monoprotic weak acid when dissolved in water is 0.36% dissociated and produces a solution with a pH of 3.10. Calculate the Ka of the acid.

I have tried a few methods. I know that, Percent Dissociation = (Amount Dissociated / Initial Concentration) x 100%

I am stuck on two ideas here, and both are wrong, but I hope I am headed in the sort of right direction.

1st attempt was; [H₃O⁺] = 10^-3.10
[H₃O⁺] = 7.9x10^-4.
I assumed the 7.9x10^-4 would be the same for the A^-;

(7.9x10^-4)(.36) = 2.84x10^-4
(7.9x10^-4)(.36) = 2.84x10^-4
(7.94x10^-4(.64) = 5.1x10^-4

Then, I set it up in Ka = Products/Reactants. The final answer was wrong.

The second method I used was the I C E chart method, but I couldn't set it up properly to fit the equation format for the life of me.

Please, can someone lead me towards the light! I would appreciate it VERY much.

EDIT: Nevermind! I literally was one step out of bounds with this problem. When setting up the ICE chart, the -x confused me on the E line for [HA]. I substituted Initial of [HA] for [HA_o] and set the Equilibrium of [HA] to [HA_o]-x. I used some of the methods above to convert the pH to Molarity and used the Percent Dissociation equation to find the initial [HA], it was all cake from there. Thanks for the considerations, though!

EDIT: For those who are curious to see the exact step I took to solve this, I will write the process.

I know I need to set up an I C E chart, because I need to find the equilibrium concentrations for each chemical.

HA		H+	A-
I	HAo	0	0
C	-x	+x	+x
E	(HAo-x)	x	x

[H⁺] = 10^-3.10 = 7.9x10^-4

x = 7.9x10^-4

0.36% = (7.9x10^-4/[HA_o])x100 = 2.19 0.219

[HA_o]=2.19 0.219

Looking back at the ICE chart, I can now plug in the numbers I have found.

[HA_o]-x = 2.19 - 7.9x10^-4 = 2.18

[HA] = 2.18
[H⁺] = 7.9x10^-4
[A^-] = 7.9x10^-4

Ka = ([H⁺][A^-])/[HA]

Ka = (7.9x10^-4)²/2.18

Ka = 2.9x10^-6

[Borek]

Percent Dissociation = [H⁺]/[H₃O] x 100%.

Huh? The safest definition is:

[tex]\alpha = \frac{[A^-]}{C_a} = \frac{[A^-]}{[HA]+[A^-]}[/tex]

[Vicktor]

That was part of my confusion, I took to the web to help me solve this problem and that is basically what I found, so I tried to apply it. My initial view to solve this problem was that of which you wrote. That is what I ended up using to solve it. I am no expert at chemistry or explaining it for that matter, sorry if I confused some people. But, it is solved! alas! Thank you for the reply, Borek. I am sure I will be asking questions again very soon.

[Borek]

Please remember such a thread can be read in the future by someone, so if there is something blatantly wrong it is better to correct it/mention it in the rest of the thread.

[Vicktor]

Thank you, I have went back and edited the end of my question to include my exact steps to solving this problem.

[AWK]

(7.9x10-4)(.36) = 2.84x10-4
(7.9x10-4)(.36) = 2.84x10-4
(7.94x10-4(.64) = 5.1x10-4

This is 0.36 % but not 36 %
Why did you use different values 7.9 or 7.94

[Vicktor]

That was one of the methods I tried in the very beginning when I initially asked for help. the .36 was substituted for 36% at the time due to the equivalence. The 7.94 was an extra digit that was added by accident, although it was the next digit in the number. I posted the whole, correct, solution at the bottom so others can have a look at the right way to do it step by step. I left my mistakes up there to show what I was dealing with initially.

[Vidya]

0.36% of the initial conc = 10^-pH
If we take initial as x
0.36% of x = 10^-3.10
(0.36 *x )/100 = 7.9*10^-4
so x should be 0.221
how are you getting initial concentration of the acid as 2.19 M
Check it

[Vicktor]

My mistake, I can't seem to find the edit button anymore. But it was supposed to be .219 NOT 2.18. I will try to fix it ASAP