guys pls help how to slove problems involving addition of HCL for ex calculate the pH of 0.10M ammonia solution calculate pH after 50ml of this soln is treated with 25ml of 0.10M HCL.Kb of ammonia=1.77×10-5 (10 to the power -5)
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According to the given question,the reaction is as follows:
NH3+H2O gives rise to NH4+OH-
Kb=[NH4+].[OH-]=1.77*10^-5
Before neutralisation,
[NH4+]=[OH-]=x
Kb=x^2=1.77*10^-5 (x^2=0.10,,,given)
x=1.33*10^-5 =[OH-]
Kw=[H+][OH-]
[H+]=Kw/[OH-]
=10^-14/1.33*10^-12
=7.51*10^-12
pH = -log(7.51*10^-12)
=11.12
On addition of 25ml of 0.1M HCl solution to 50 ml of 0.1M ammonia solution, 2.5 mol of ammonia are neutralised.The resulting 75 ml solution contains the remaining unneutralised 2.5mol of NH3 molecules and 2.5 mol of NH4+
Initially NH3 & HCl are in number of mol = 2.5
At equilibrium products are formed (NH4+ & Cl- ion)
So at equilibrium ,Number of mols =2.5
75 ml of the solution contains 2.5 mol of NH4+ ions and 2.5 mol of unneutralised NH3 molecules.This NH3 exists in the following equilibrium
NH4OH ---------> NH4+ + OH -
0.033 M y y
where Y=[OH-]=[NH4+]
The final 75 ml solution after neutralization already contains 2.5 mol NH4+ ions,thus total concentration of NH4+ ions is given as:
[NH4+]=0.033+y
y is small,y is neglected
[NH4OH]=0.033M and [NH4+] = 0.033M
Kb = [NH4+] [OH-] = y(0.033) = 1.77*10^-5
thus y=1.77*10^-5 =
[OH-]=0.56*10^-9
[H+]=10^-14 = 1.77*10^-5
pH=9.24
Hope it helps!!!!