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.

Question

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

Answer 1

Would you show your work and let me check it? It's the Henderson-Hasselbalch equation.

Answer 2

I keep getting about 1.08M for NH4Cl.

Answer 3

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

Answer 4

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)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)

.5556/.600M=acid
acid=.925925...M
molarity (.925925)*2.10L=

Answer 5

yes you are totally right.
Thank you so much!!!!!!!