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Somebody please answer this pH of solution containing 0.25M ammonia and 0.44M ammonium chloride solution is0.25 Kb NH3=1.4*10^-5

Somebody please answer this 
pH of solution containing 0.25M ammonia and 0.44M ammonium chloride solution is0.25 
Kb NH3=1.4*10^-5

Grade:12

1 Answers

Sahani Kumar
99 Points
5 years ago
Given:  [salt(NH4Cl)]=0.44
             [base(NH3)]=0.25.     Kb=1.4*10^-5
Now,  pOH= – logKb+log[salt]/[Base] 
           pOH= – log(1.4*10^-5)+log[0.44]/[0.25]
           pOH= (– log1.4)+ (– log10^-5)+(log1.76)
           pOH=( –0.146)+(5)+0.245
           pOH=4.901
Now we know that pH=pKw – pOH.  where pKw=14
   pH= 14 – 4.901
   pOH=9.099

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