Determine the ammonia concentration of an aqueous solution that has a ph of 11.50. the equation for the dissociation of nh3 (k b = 1.8 ã 10-5) is nh3(aq) + h2o(l) â nh4 +(aq) + oh-(aq).

When PH = 11.5 so, we can get the POH from this formula:

PH + POH = 14 

11.5 + POH = 14 

∴POH = 2.5 

and we can get [OH] from this formula:

POH = -㏒[OH-]

2.5 = -㏒[OH-]

∴ [OH] =  0.0032

according to the reaction equation:

NH3(aq) + H2O(l) ↔ NH4+ (aq)   +   OH-(aq)

so, Kb = [NH4+][OH] / [NH3]

when [NH4+] = [OH]

by substitution with Kb & [OH] 

1.8 x 10^-5 =  (0.0032)^2 / [NH3]

∴ [NH3] = 0.569 M

∴ the concentration of ammonia = 0.569 M

Ibelieve your answer is true