Question

I need help with this question. I need to calculate the volume amount of the reagent needed which in my case is ammonium chloride ph: 8.75 . We need to prepare it as a 0.100 M aqueous solution to prepare a 70 ml sample of this buffer.

Stations 13 to 16: Ammonium chloride buffer, pH 8.75 Stations 17 to 20: Ammonium chloride buffer, pH 9.25 Stations 21 to 24: Ammonium chloride buffer, pH 9.75 or the assigned buffer.This is your pre-lah assignment. It can be hand-written. If the measured pH is different from the assigned pH by more than 0.09 pH units, you need to adiust it by adding acid Gf pH is high) or base (if pH is low) and correct the volume

Answer 1

total molarity= 0.1 M

total V = 70 mL

total mol = mmol = MV = 70*0.1 = 7 mmol

this is;

NH3 + NH4+ = 7

from buffer equation

pH = pKa + log(NH3/NH4+)

pKa for NH3 = 9.25

8.75 = 9.25 + log(NH3/NH4+)

(NH3/NH4+) = 10^(8.75 -9.25 ) = 0.3162

NH3 = 0.3162*NH4+

NH3 + NH4+ = 7

0.3162*NH4+ + NH4+ = 7

1.3162*NH4+ = 7

NH4+ = 7/1.3162 = 5.31834 mmol

probably, you will have NH4Cl souirce of NH4+

mass = mmol*MW = 5.31834*53.4915 = 284.48 mg of NH4Cl required

(NH3 = 0.3162*5.31834 = 1.6816 mmol

this is most likely from a solution so...

if Molarity of NH3 source is 0.1M; then add:

V = mmol/M = (1.6816)/(0.1) = 16.86 mL of NH3 at 0.1 M

finally, ad dup water up to V = 70 mL