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
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