Nh3 is a weak base (kb = 1.8 × 10–5 m) and so the salt nh4cl acts as a weak acid. what is the ph of a solution that is 0.013 m in nh4cl?

First, we need to get the value of Ka:

when Ka = Kw / Kb 

we have Kb = 1.8 x 10^-5 

and Kw = 3.99 x 10^-16  so, by substitution:

Ka = (3.99 x 10^-16) /  (1.8 x 10^-5) = 2.2 x 10^-11

by using the ICE table :

                 NH4+  + H2O →NH3 +  H+
intial          0.013                       0         0 

change       -X                          +X      +X

Equ        (0.013-X)                      X         X

when Ka = [NH3][H+] / [NH4+] 

by substitution:

2.2 x 10^-11 = X^2 / (0.013 - X)  by solving this equation for X

∴X = 5.35 x 10^-7

∴[H+] = X = 5.35 x 10^-7

∴PH = - ㏒[H+]
 
        = -㏒(5.35 x 10^-7)
        = 6.27

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