Buffer Problem.....?

Question

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 650. mL of 0.894-M solution of NH3 in order to prepare a pH = 9.00 buffer?

I keep getting 17.48 but it is the wrong answer.
This is how I am working it out.
PH=pKa+log(b/a)

9=9.25+log(x/.5811)
x=.326777 mol
Then I multiply by Mm to get 17.48 grams. I am doing something wrong and don't know what it is.

Answer 1

You've put the NH3 on the bottom instead of on the top of the log term. NH4+ is the acid, and NH3 is the base.

Try it that way and see what happens...

Answer 2

sorry if i am wrong.i did this buffer topic a year and a half ago in grade 11 but now i have joined medical field thats why i may go wrong though i will try to help you out:
i dont know whether the calculations which u have mentioned have been done right or wrong.let me consider it right,still there is a mistake.the "x" which u have found is mol/litre(concentration),not mol.
now here is how to do it further:
to 1 litre,the amount of x added = .326777mol
to .65 litre,the amount of x added will be=.326777 X .65=.212 mol.
Now for changing moles into weight in grams,multiply with Mm:
.212 X 53.49= 11.36g.(answer)
is it right or not?

Answer 3

Lancenigo di Villorba (TV), Italy

If I am correct, you have to calculate the required amount of Ammoniac Salt (e.g. Ammonium Chloride, NH4Cl) in order to formulate CLASSICAL AMMONIA AQUEOUS BUFFER.
So, I will refer my calculations to assigned volume as 650 mL.
Since your desired pH's level is 9.0, I calculate HYDROXYL ION's concentration

|OH-| = ALOG(pH - 14) = 1.0E-5 M

a datum involved in the HENDERSON-HASSELBACH's EQUATION

1.8 = 1.8E-5 / 1.0E-5 = Kb / |OH-| = |NH4+| / |NH3|

In the same liquid mass, the Molarity's Ratio results equal to Mole's Ratio

1.8 = |NH4+| / |NH3| = (NH4Cl moles) / (NH3 moles)
(NH3 moles) = (NH3 molarity) * Volume = 0.894 * 0.65 = 0.58
(NH4Cl moles) = (NH4Cl mass) / (NH4Cl M.W.)
(NH4Cl moles) = (NH4Cl mass) / 53.45

hence you obtain

(NH4Cl mass) = 1.8 * 0.58 * 53.45 = 55.9 g

I hope you appreciate it.