Analytical chemistry

What is a pH of 20 mL of 0.08 M NH3 added to 40 mL of 0.04 M HCl?

NH3 + HCl = NH4+ + Cl-

c(NH3)=(0.02L * 0.08M)/0.06L= 0.02667 M
c(HCl)=(0.04L * 0.04M)/0.06L= 0.02667 M
c(NH4+)=c(HCl)

Kb=[NH4+][OH-]/[NH3]
Kb=1.76*10^(-5)

So i tried to calculate it like this:
Kb=[OH-]
pOH=4.7545
pH=9.2455
But i know this is the wrong way, i just don't understand problems with buffers that well.

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asked by Gloria
  1. But your problem with this problem is this isn't a buffer. Since you have the same molarity for HCl as for NH3, then you have NH4Cl (the salt) BUT no acid or base to go with it to make a buffered solution. What you have instead is the salt (by itself) in water so this is a hydrolysis problem and the pH is determined by the hydrolysis of the NH4^+.
    ..........NH4^+ + H2O ==> H3O^+ NH3
    I.......0.02667............0.....0
    C..........-x..............x.....x
    E.......0.02667-x..........x.....x

    Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.02667-x) and solve for x then convert to pH.

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    posted by DrBob222

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