[benworld128]

How many grams of dry Nh4CL need to be added to 2.40 L of a 0.300  M solution of ammonia, Nh3, to prepare a buffer solution that has a pH of 8.81? Kb for ammonia is 1.8 X 10^{-5}.

My Way:
--------
pKb = 4.7
pOH = 5.15
pKb = pOH + log [NH4+]/ [NH3]
4.7 - 5.15 = - 0.45
10^-0.45=0.355
0.355 * 0.300 = 0.1065
 
[NH4+]= 0.1065
moles = 0.1065 x 2.40 L= 0.2556
Mass = 0.2556 x 53.49 g/mol= 13.67 g

is this right ?

because I the mastering chem says its wrong.

Plz help.

[UG]

pOH = 5.15

How did you get this value?

[benworld128]

sorry its 5.19

[UG]

pKb = pOH + log [NH4+]/ [NH3]

The formula should be pOH = pKb + log([NH₄⁺]/ [NH₃])

[benworld128]

Okay since given Ph I will solve for the ratio

ph = pka + log(base/acid)

8.81 = 9.26 + log(base/acid)
-0.41 = log(base/acid)
10^-0.41 = base / acid

base / acid =  0.355

solving for acid = 0.355/0.300 = 1.18

1.18 * 0.24 L =  0.2832 mol * 53.49 g  = 15.14 g

is that correct ?

[UG]

base / acid =  0.355

solving for acid = 0.355/0.300 = 1.18

1.18 * 0.24 L =  0.2832 mol * 53.49 g  = 15.14 g

is that correct ?

Solving for acid should be the other way round, 0.300/0.355 = 0.845
The volume should be 2.4 L instead of 0.24 L

[benworld128]

base / acid =  0.355

solving for acid = 0.355/0.300 = 1.18

1.18 * 0.24 L =  0.2832 mol * 53.49 g  = 15.14 g

is that correct ?

Solving for acid should be the other way round, 0.300/0.355 = 0.845
The volume should be 2.4 L instead of 0.24 L

I don't know what I'm doing. Can someone guide me in right direction at least ?

[UG]

You are nearly there, you have worked out the concentration of acid needed, which was calculated by 0.300/0.355 = 0.845 M.
Then to work out the number of moles of ammonium chloride, you go 0.845 M * 2.4 L = 2.028 mol. Then multiply by the molar mass to get 2.028*53.49 = 108.5 g