Asked by Jessica

How many grams of dry NH4Cl need to be added to 2.10L of a 0.600M solution of ammonia,NH3 , to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8x10^-5.

I just wanted to make sure I dd this write, can you please check my work?
Thank you!!!!

I found the pKa which was 9.255..
then I used that and plugged it in the Hasselbalch (i think that's what it's called) equation.
I got .92592....M for NH4Cl and i used that Molarity and multiplied it by the liters.
so the mass was 1.94 moles of NH4Cl
I then used that to find the grams and my answer came out to be 104.02 grams of NH4Cl
Answered by DrBob222
Would you show your work and let me check it? It's the Henderson-Hasselbalch equation.
Answered by DrBob222
I keep getting about 1.08M for NH4Cl.
Answered by Jessica
I got 9.25527 for the pKa
then for the Henderson-Hasselbalch equation:
9.00=9.25527+log(.600/acid)
10^(-.25527)=(.600/acid)
.5556/.600M=acid
acid=.925925...M
molarity (.925925)*2.10L=

1.94 moles of NH4Cl
1.94 moles *53.4 grams=104.02g of NH4Cl
Answered by DrBob222
I think you made an algebra error from line 4 to line 5 in your work.

I got 9.25527 for the pKa
then for the Henderson-Hasselbalch equation:
9.00=9.25527+log(.600/acid)
10^(-.25527)=(.600/acid)<b>correct to this line. Then
0.5556 = 0.6/acid and
acid = 0.6/0.5556 = 1.08 which is the reciprocal of your number)</b>

.5556/.600M=acid
acid=.925925...M
molarity (.925925)*2.10L=
Answered by Jessica (DrBob222 thank you so much!!!!)
yes you are totally right.
Thank you so much!!!!!!!
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