Question

What is the pH of a buffer system that contains 0.200M hydrocyanic acid (HCN) and 0.230M sodium cyanide (NaCN)? The pKa of hydrocyanic acid is 9.31.

Answer 1

I have noo clue what this is even talking about! ):
Please help!

Answer 2

Use the Henderson-Hasselbalch equation pH=pKa+log([Base]/[Acid]).
You have the concentrations of both the weak acid (HCN) and it's conjugate base, and you also have the pKa. And all you do is plug in. =]

Answer 3

Thank you for replying!!
Ok so do I have to minus anything from the concentrations given?
Cause my teacher did that in class and that's the part that I got stuck on!!

Answer 4

And Also do I dive the base and the acid and thennn do i do the log of it?

Answer 5

Nope, not for this problem since you're not adding any more of acid or base to it. It's just a plain buffer solution with nothing added to it and they're just asking for the pH of it. Nothing changes. Aaand yes, divide the base concentration with the acid conc. first and then do log and then add that value onto the pKa. And that's all =]

Answer 6

It is the neg log or just the reg log? I have a minus sign written off before the long and no plus sign.. maybe i goofed :/

Answer 7

You make it super simple, thanks!! there's like 5 of these and I was stuck on this one!
:S

Answer 8

Just regular log, hmm what equation are you using? The one I used has no negatives in it. You may or may not get a negative value for log([base]/[acid]), but you still will adding that value to pKa. And no honestly I was confused with buffer problems at first too like you don't even know but it's like, you just get better with practice and eventually you'll just breeze through it lol. Practice makes perfect =]

Answer 9

It's not accepting my answer): Did I do something wrong? I divided the acid and the base and then did the reg log which gave me a neg answer.. and I added that to 9.31 and it says it's wrong! ):

Answer 10

Here I'll work through it with you so we can check if we get the same answer.

Okay so... pH=pKa+log([base]/[acid])

pH=9.31+(log(0.230 M/0.200 M))

pH=9.31+0.0606978404

pH=9.3706, rounded to 3sig figs would get you pH of 9.37.

Answer 11

Are you using masteringchem?

Answer 12

Thank you sooo much, youuu are amazing!! and yesss I am.. unfortunately! :/ I'm guessing you have experience! hahaa.. it's sooo picky!

Answer 13

oh!!! and I caught my mistake i did the acid over the baseee because i have that in my notes but.. maybe you should be our teacher! hahaha.

Answer 14

Really? Scary how your teacher told you the wrong equation, I'm sorry =[
But always, always use this equation whenever doing ANY sort of buffer problems
Here it is again: pH=pKa+log([Base]/[Acid])

Answer 15

Well now I have 3 more to go so i'm sure this will help out a lot!
Thankk you zoo much, and btw your dog is adorbz! haha(:
I have highlighted the new formula you gave me in my notes.
And again, thanks a million!!(:

Answer 16

soo*

Answer 17

Lol thanks! and good luck on your problems nd also lmk if you still need help. I know the ones where there's like acid or base added get more confusing, so just ask and i'll see if i can help =]