Determine the ph of a 0.188 m nh3 solution at 25°c. the kb of nh3 is 1.76 × 10-5.

When NH3 is dissolved in water, it dissociates  partially producing NH4+ ions and OH- ions. It has an equation:
NH3 + H2O → NH4+ + OH- 

We use the Kb expression to determine the [OH-] concentration,

Kb = [NH4+] [OH-] /* [NH3] 

We can write NH4+ as OH- since they are of equal ratio. 
(1.76*10^-5) = [OH-]² / 0.188 
[OH-]² = 3.3088*10^-6 
[OH-] = 1.819*10^-3 

We calculate for H+ concentration as follows:

[H+] [OH-] = 10^-14 
[H+] = 10^-14 / [OH-] 
[H+] = 10^-14 / (1.819*10^-3) 
[H+] = 5.50*10^-12 

pH = -log [H+] 
pH = -log (5.5*10^-12) 
pH = 11.26

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