[webassignbuddy]

How many grams (to the nearest 0.01 g) of NH₄Cl (Mm = 53.49 g/mol) must be added to 650. mL of 1.345-M solution of NH₃ in order to prepare a pH = 9.40 buffer?

1. 650 mL x (1.345 mmol/mL) = 874.25 mmol NH₃

2. NH₄Cl would be considered H₃O⁺ right?

3. Reaction Table

H₃O⁺      +     NH₃           H₂O    +    NH₄⁺
    x             874.25                                0           mmol
   -x                 -x                                 +x          mmol
    0            874.25-x                              x           mmol

4. pH = pK_a + log (n_b/n_a)

pKa of NH4+ is 9.25

9.40 = 9.25 + log(874.25 - x / x)
0.15 = log(874.25 - x / x)
10^0.15 = (874.25 - x / x)
1.4125 = (874.25 - x / x)
1.4125x = 874.25 - x
1.4125x + x  = 874.25
2.4125x = 874.25
x = 874.25/2.4125
x = 362.3 mmol of NH₄Cl required
x = 0.3623 mol of NH₄Cl required

4. Convert to grams

0.3623 mol x (53.49 g/mol) = 19.379 g = 19.40 g

Is that correct?

[Borek]

http://www.chembuddy.com/?left=buffers&right=composition-calculation

2. NH4Cl would be considered H3O+ right?

No, NH₄⁺ is NH₄⁺, it is not equivalent to the concentration of H⁺.

Then your ICE table is wrong, as it doesn't describe the solution to which solid NH₄Cl was added, but one containing initially ammonia and strong acid only.

No wonder the final answer is off as well.

[webassignbuddy]

http://www.chembuddy.com/?left=buffers&right=composition-calculation

2. NH4Cl would be considered H3O+ right?

No, NH₄⁺ is NH₄⁺, it is not equivalent to the concentration of H⁺.

Then your ICE table is wrong, as it doesn't describe the solution to which solid NH₄Cl was added, but one containing initially ammonia and strong acid only.

No wonder the final answer is off as well.

Oh ok.
I got the answer!
Thanks!