[starter]

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 initial moles and 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.

However i have a feeling i need the Kb value. Help please

[Borek]

then i take the molar mass of ammonia which is 17 and i divide the .102/17=.006

That's the thing I have never seen before... Molar concentration divided by molar mass. No idea what you are trying to achieve here.

At least you are right that you should use Kb. See http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

[starter]

The way i approach this question is as follows i have the initial moles and volume so M1V1=M2V2 ...
so 25*.204= 5.1
then 5.1/50= .102
then i would use my kb value which is 1.8x10^-5 and i would multiple it by the .102
then i do -log 1.836x10^-6 which then gives me 5.736
so then i subtract that from 14-5.736= 8.26
I think thats right how does that look Borek

[starter]

my bad when i get 1.8x10^-5 i need to squareroot that and then that number will be the number i take away from 14 and my final answer should be 11.13.

[Borek]

OK now. Note that you have to know whether you can use simplified equation or not - it holds only if the assumption done during derivation holds.