Question

calculate the amount of (NH4)2 SO4 which must be added to 500 ml of 0.2 (M) NH3 to yield a solution of pH= 9.35 pKb (NH4OH) = 4.74

Answer 1

 Best Answer

NH3 + H2O --> NH4+ + OH- 
pKb NH3 = 4.7 
pOH = 14 - 9.35 = 4.65 
for pOH = 4.65, [OH-] = 2.24x10^-5M 

KbNH3 = 1.82x10^-5 = [NH4+][OH-]/[NH3] 
1.82x10^-5 = [NH4+][2.24x10^-5] / 0.2 
[NH4+] = 0.162M 

0.162M x (1(NH4)2SO4 / 2NH4+) = 0.0812M (NH4)2SO4 
0.0812M x 0.5L = 0.0406moles (NH4)2SO4 
0.0406moles x 132g/mole = 5.36g required

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