24thanx sir.......
An ammonia ammonium chloride buffer has a pH value of 9 with[NH3]-0.25.By how much will the pH will change if 75 mL 0.1M KOH be added to 200 mL buffer solution.Kb=2*10-5
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2 Answers
1357pH=9 so [OH-]=10-5
NH3+H2O [eq] NH4++OH-
0.25 x 10-5
x*10-5/0.25=Kb=2*10-5
x=0.5
NH3 + H2O [eq] NH4+ + OH-
mmoles 0.25*200 0.5*200 10-5*200+0.1*75~7.5
let the whole of OH- is consumed
NH3 + H2O [eq] NH4+ + OH-
mmoles 50+7.5 100-7.5 0
mmoles 57.5-x 92.5+x x
total volume = V=275 ml
(x/275)((92.5+x)/275)/((57.5-x)/275)=Kb
let x be very small
x*92.5/57.5=275Kb
x=3.42*10-3
[OH-]=x/275=1.24*10-5M
pOH=4.9 and pH=9.1
so pH changes by 0.1