Achiever 1 genchem question

[iwantdent]

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What is [H+) of 0.20 M hydrocyanic acid solution that has been made 0.40 M of sodium cyanide NaCN? (Assume ionization constant for hydrocyanic acid is Ka= 4.0x10^-10.)

I am not sure if the wording of the question is weird. But any help would be appreciated! Thanks!

 

[ballyhoo]

I think you use hendersson-hasselbach!

 

[mario0923]

What is [H+) of 0.20 M hydrocyanic acid solution that has been made 0.40 M of sodium cyanide NaCN? (Assume ionization constant for hydrocyanic acid is Ka= 4.0x10^-10.)

I am not sure if the wording of the question is weird. But any help would be appreciated! Thanks!

this is pH of salt
just use the Chad's equation

If you want here is the link
http://forums.studentdoctor.net/showthread.php?t=926308
Destroyer had the same question and somebody asked for it and I answered it her/him.

 

[Ari Rezaei]

I don't think its just the pH of a salt since acid was given too along with a conjugate base. I didn't think about using henderson hasselbach. I think first you set up an ICE table and just solve for X since its only 1 mol of H+.

Ka = [H+]*[CN-] / [HCN]
But we are given the intials and need the equilibriums. So:

Ka = [x]*[0.40+x] / [0.20-x]
I think this is where you can just drop the x's in the 0.40+x and 0.20-x to make the problem easier (because x is so small it becomes insignificant)


Ka = [x]*[0.40] / [0.20]

Solve for x
x = 2.0e-10

And x is our final answer because there is only 1 mol of H+.

 

[Ari Rezaei]

You could also use henderson hasselbach, just tried it out. But good luck without a calculator 🙂. Also requires you to go from -log to [H+] back and forth a couple times and find the log of 2.

 

[Stacker]

.

 

[mario0923]

I don't think its just the pH of a salt since acid was given too along with a conjugate base. I didn't think about using henderson hasselbach. I think first you set up an ICE table and just solve for X since its only 1 mol of H+.

Ka = [H+]*[CN-] / [HCN]
But we are given the intials and need the equilibriums. So:

Ka = [x]*[0.40+x] / [0.20-x]
I think this is where you can just drop the x's in the 0.40+x and 0.20-x to make the problem easier (because x is so small it becomes insignificant)


Ka = [x]*[0.40] / [0.20]

Solve for x
x = 2.0e-10

And x is our final answer because there is only 1 mol of H+.

well...I think I didn't read carefully sorry if I was wrong!