Can somebody check my ph question ?

Question

well the question goes like this :Calculate the pH of an ammonia solution made by diluting 25.0 mL of 0.204 M NH3 to a total volume of 50.0 mL.
a.) 2.87 b.) 8.26 c.) 9.25 d.) 11.13 e.) 11.78

The way i approach this question is as follows i have the intial mole sand volume so M1V1=M2V2 ...
so 25*.204= 5.1
then 5.1/50= .102
then i take the molar mass of ammonia which is 17 and i divide the .102/17=.006
then i do -log (.006) =2.22
so then i do 14-2.22= 11.78 which is one of the answers if anybody can confirm or deny, any additional help is appreciated

Gerald you got me confused now how do i find the Kb and is the rest seem reasonable

Answer 1

The final conc of NH4OH is 0.102 M (we agree)
Kb for NH4OH = 1.8x10^-5
After dissociation [NH4+] = [OH-] and [NH4OH] = 0.102 - [OH-] which is about 0.102
1.8x10^-5 = [OH-][OH-] /(0.102)
let X = [OH-]
1.8x10^-5 (0.102) = X^2
1.836x10^-6 = X^2
X = 0.00135
pOH = 2.87
pH = 14 - 2.87 = 11.13

Answer 2

Your calculation completely ignores the fact that ammonia is a weak alkali, and is very little dissociated in solution.
You need to use Kb as well as the molarity.