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`        pH of a solution containing 0.25 M ammonia and 0.4M ammonium chloride solution given kb of NH3=1.8*10^-5`
one year ago

## Answers : (1)

Sahani Kumar
99 Points
```							Given:  [salt(NH4Cl)]=0.44             [base(NH3)]=0.25.     Kb=1.4*10^-5Now,  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=4.901Now we know that pH=pKw – pOH.  where pKw=14   pH= 14 – 4.901   pOH=9.099
```
one year ago
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