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What volume of 0.40 M NH3 solution must be added to 1Lof 0.1 m NH4Cl solution to give a buffer having pH of 10?For NH3, KB=1.8^10-5.
[H+] = 10-10 [OH-] = Kw/H+ = 10-4 moles of NH4Cl = 0.1*1=0.1 for V Litre of NH3 ,moles of NH3 = 0.4*V now , Kb = [salt][OH-]/[base] 1.8*10-5 = (moles of salt )*10-4/(moles of base) moles of base(NH3) = 10/1.8*4=1.38L
[H+] = 10-10
[OH-] = Kw/H+ = 10-4
moles of NH4Cl = 0.1*1=0.1
for V Litre of NH3 ,moles of NH3 = 0.4*V
now ,
Kb = [salt][OH-]/[base]
1.8*10-5 = (moles of salt )*10-4/(moles of base)
moles of base(NH3) = 10/1.8*4=1.38L
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