× Didn't find what you were looking for? Ask a question
Top Posters
Since Sunday
5
a
5
k
5
c
5
B
5
l
5
C
4
s
4
a
4
t
4
i
4
r
4
Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
New Topic  
ripsmithy ripsmithy
wrote...
Posts: 19
Rep: 0 0
11 years ago
What mass of ammonium chloride should be added to 2.45 L of a 0.150 M {\rm{NH}}_3 in order to obtain a buffer with a pH of 9.65?
I think I am supposed to use the Henderson-Hasselbalch equation but I can't figure it out! Please help!
Read 317 times
2 Replies

Related Topics

Replies
wrote...
#1
11 years ago
Keep calm

Yes, use Henderson-Hasselbach

Work out the ratio of [NH3] to [NH4+] that you need.

You know [NH3], so what must [NH4+] be?

That's moles/L; you need g/2.45L. I think you know how to do that, now you can see where you're going.

Good luck
wrote...
#2
11 years ago Edited: 11 years ago, bio_man
You are correct in that you must use the Hendersom - Hasselbalch equation.
pOH  = pKb + log ([HB+]/[ B])
Make a list of things you know or can get easily:
 pH = 9.65 pOH = 4.35
pKb ( NH3)  = 4.75
  = 0.150

pKb for NH3 is 4.75, you want pOH = 4.35.
Therefore:  log ([HB+]/[ B]) = -0.4
[HB+] / [ B] = 10^-0.4
[HB+] / 0.150 = 0.398
[HB] = 0.150*0.398
[HB+] = 0.0597
The molarity of the NH4Cl must be 0.0597M

Let us check our calculation:
 pOH = pKa + log (0.0597/0.15)
pOH = 4.75 + log 0.398
pOH = 4.75 + ( -0.4)
pOH = 4.35
pH = 14.00-4.35 = 9.65 It all works out

All we need is the mass of KCl to produce 2.45L of 0.0597M solution
 MOlar mass NH4Cl = 53.4917 g/mol
Mass required = 54.4917*2.45*0.0597 = 7.97g NH4Cl required.
New Topic      
Quick Reply
Explore
Post your homework questions and get free online help from our incredible volunteers
  1272 People Browsing
Start New Topic Take the Tour CoursesNew Study Tools Topics Trending Browse by Textbook
Live Chat
Related Images
  
 116
  
 32
  
 286
Your Opinion
Do you believe in global warming?
Votes: 370
Previous poll results: Which country would you like to visit for its food?