Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: BlondeGirl2012 on August 14, 2012, 05:49:10 PM

You have 200 mL of 1 M ammonia solution (pKa=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.25? Answer: 100ml
Guys please help me. :(
I know how to do the question when there is 100 ml of 1m ammonia solution. But now the question is 200 mL and I am getting confused. Can someone please show me the steps? Thank you
This is the solution for when it's 100 ml of 1m ammonia. I looked at it and I did myself and got it right.
"So we know that from the HH equation, 9.5 = 9.25 + log (NH3)/(NH4+)
Therefore log (NH3/NH4) = 0.25, use a calculator to find that log x =0.25, where x = 1.78, the ratio of NH3/NH4+.
NH3 ratio to start is 0.1, so basically since HCl completely disassociates (and in a 1 H+ to HCl manner) we know that any HCl added will form the conjugate acid NH4+. If we represent the amount of HCl added as x, we know that NH3 will be 0.1  x at the buffer solution (starting ammonia will react with the HCl added). We know that x will also be the amount of NH4+ formed.
Therefore we can represent NH3/NH4+ as (0.1x)/x, we calculated this ratio as 1.78 earlier. Simply solve for x to get the amount of HCl added (also the amount of NH4+ formed)."

Yes take HH equation. Write it down.
What is the equation between HCl and NH_{3}?
And special case pKS = pH

Interesting thing about your problem is that the solution you listed doesn't mention volume at all  yet it is volume being 200 mL and not 100 mL that confuses you. Perhaps it doesn't matter till the end of the problem, when all you need to do is the volume/concentration/number of moles thing?

Hi guys. Thank you for your comments. If any of you can actually solve it and get 100ml as the final answer, can you please show the steps?
This is what I did when there is 100 ml of 1M ammonia solution (pKA=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.25?
NH3 + HOH <==> NH4+ + OH
9.5=9.25 + log (NH3/NH4+)
log(NH3/NH4+)= 1.78
NH3 + HCl = NH4Cl
HCL completely dissociates
Any HCL added will form the conjugate acid NH4+, so NH4+ = amount of HCL added
NH3/NH4+ =1.78 = (0.1  x)/(x)
x=0.357=36 ml

Hi guys. Thank you for your comments. If any of you can actually solve it and get 100ml as the final answer, can you please show the steps?
This is what I did when there is 100 ml of 1M ammonia solution (pKA=9.25). What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.25?
NH3 + HOH <==> NH4+ + OH
9.5=9.25 + log (NH3/NH4+)
log(NH3/NH4+)= 1.78
NH3 + HCl = NH4Cl
HCL completely dissociates
Any HCL added will form the conjugate acid NH4+, so NH4+ = amount of HCL added
NH3/NH4+ =1.78 = (0.1  x)/(x)
x=0.357=36 ml
Where has the 9.5 come from?

and the 0.1 is also wrong

"What volume of 1 M hydrochloric acid is needed to prepare buffer with pH=9.5" not ph=9.25
Thanks.

Think again your approach. In the HH equation you have concentration not mole.
Develop again the equation.

Hunter2, thanks for your reply.
This is the equation if I use concentration,
(1x)/(x)=1.78
1x=1.78x
1=2.78x
x=0.359=36 ml. I got it right when it's 100 ml of 1M ammonia solution.
But could you kindly show the steps for getting answer=100 ml when there is 200 ml of 1M ammonia solution?
According to HH equation, as you said, if I use concentration, it's still
(1x)/(x)=1.78 when there is 200 ml of 1M ammonia solution.

The x is 0.359 mol/l and for 100 ml it guides you to the 36 ml.
For 200 ml you need the double amount 71,8 ml

The problem is, the answer is 100 ml not 71.8 ml. That's why I am curious how they got the answer 100 ml. ??? And that's why I am asking on a forum.

100 mL is a correct answer to the initial question (pH=pKa=9.25).