solubility problem HCN

[andyjl]

Advertisement - Members don't see this ad

Find the Ka of HCN given that a 0.20 M solution of HCN(aq) is 0.002% ionized at 25 degrees celcius

 

[laundry]

So since it's .002% ionized, only that percentage of the molarity will represent the ion concentration, and since that's an insanely small number we can leave the Ka denominator at 0.2.

0.002% of 0.2 can be derived easily by doing this:

(2/10)(2/1000)=(4/10000)=1/2500

Thus, our equation looks:

Ka = (H+)(CN-)/(HCN)= ((1/2500)^2 )/ 0.2

Again, you can rewrite that as:

(1/2500^2)/(2/10) = 5/ (2500^2) Now...since 25^2 is 625 (you should just know that), we get:

5/6250000 = 1/1250000 again, we know that 1000/125 is 8, so the answer is:
Ka= 8.0 x 10^-7

I'm at the American Society of Mammalogist and I've just unpacked about 5 min ago, so forgive me if I'm of by a 10 fold or something, but you should now know how to do this problem regardless of the math... =)

Cheers,

P.S. I hope this is correct, someone please double check...👍

 

[PD3]

I got the same thing, but looked at the amount ionized differently. It makes sense to me to think of the percent ionized as .00002=(X/.02) where X is the amount that is ionized. This gives you the 1/2500 that you spoke of. Other than that I did it the same. This just makes more sense to me.

 

[joonkimdds]

So since it's .002% ionized, only that percentage of the molarity will represent the ion concentration, and since that's an insanely small number we can leave the Ka denominator at 0.2.

0.002% of 0.2 can be derived easily by doing this:

(2/10)(2/1000)=(4/10000)=1/2500

0.002% = (x/0.2) *100%
x = (0.002%/100%)*0.2 = 4*10^-6

1/2500 = 4*10^-4 so I think you didn't divide by 100????????


and if I keep continue this
Ka = [(4*10^-6)^2]/0.2 = 8 x 10^-11