Alright, here is my question:
22. References list the Ka value of hydrosulphuric acid, H 2S(aq) , as
1.1 x 10^–7 at 25°C. Assume a solution is prepared by dissolving 1.00 g of H2S(g) to make 1.00 L of acidic solution. What percentage of this solution will ionize?
And I don't know how to find the percent which would ionize, but I have figured out this information:
Concentration is equal to 0.029334mol/L
[H+] is equal to 5.68x10^-5
Can you help me on this question, thanks!
I concur with 0.0293 M H2S as well as (H^+)=5.68 x 10^-5 M.
So %ionized = [(amt ionized)/0.0293]x100 = (5.68 x 10^-5/0.0293)x 100 = ??
I prefer to call 1.1 x 10^-7 k1 and not Ka. I hope this helps. %ionization is nothing more than the fractional part of the initial amount that ended up as ions (then converted to percent by multiplying by 100).
To find the percentage of the solution that will ionize, you need to calculate the amount of hydrosulfuric acid that will ionize and then divide it by the initial concentration of the acid.
First, let's calculate the amount of H2S that will ionize. The initial concentration of H2S is 0.0293 mol/L, as you mentioned. The hydrosulfuric acid ionizes according to the following equation:
H2S ⇌ H+ + HS-
From the equation, we can see that for every H2S molecule that ionizes, we get one H+ ion and one HS- ion. Therefore, the amount of H2S that will ionize is equal to the concentration of H+ ions.
You provided the concentration of H+ ions, which is 5.68 x 10^-5 mol/L. So, the amount of H2S that ionizes is also 5.68 x 10^-5 mol/L.
Now, we can calculate the percentage of the solution that will ionize. We need to divide the amount of ionized H2S by the initial concentration of H2S and multiply by 100:
% ionized = (amount ionized / initial concentration) x 100
= (5.68 x 10^-5 mol/L / 0.0293 mol/L) x 100
= 0.194 x 100
= 19.4%
Therefore, approximately 19.4% of the hydrosulfuric acid solution will ionize.