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`        10ml of 0.5M RNH3Cl (kh = 10^-9)+40ml of0.125M KOHFind pH.`
2 years ago

Rajat Jain
11 Points
```							RNH3Cl+KOH--->RNH3OH+KCl Millimoles:initial:5.   5.            0.              0.                  Final: 0.    0.           5.               5Kh for RNH3Cl is as good as Ka(acid dissociation constant) for RNH3+.So Kb for RNH3OH will be Kw(Ionic product of water) divided by Ka,since RNH3OH and RNH3+ are conjugate acid-base pair. Now,we have Kb=10^-5So,.                  RNH3OH --->RNH3+. + OH-Initial Concentration: 5/50.       0.               0Final.               5(1-x)/50     5x./50         5x/50Therefore,. Kb=(5x)(5x)/{5(1-x)*50} :.             10^-5=x^2/10(1-x)Assuming x can be neglected in front of 1 We get x=0.01,which justifies our assumption. Now, [OH-]=5(0.01)/50=10^-3Therefore, pOH=3 and pH=11.Note:Here,contribution of water in concentration of OH- was neglected as [Base]Kb ≥100Kw !!Note-2: In cases,where degree of dissociation of base is negligible and [Base]Kb≥100Kw,[OH-] can be directly  written as ([Base]Kb)^0.5!!
```
2 years ago
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