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ripsmithy ripsmithy
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12 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!
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#1
12 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
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#2
12 years ago Edited: 12 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.
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