well dont we have to multiply the [.08943] by the (.025L) then divide by .1 then multiply by 1000ml = to give the amount of ml need to reach the equ.pt. which cam out to be 22.357

then

do the

Initial = .08943 0 0

Equ = .08943-x x x

x2= Kb of ammonia * .08943M

x = square root of Kb * .08943M

which gives us the [OH] CONCENTRATION

take the -log(of the[OH]

which is the pOH ?

then just take 14- pOH

which gives pH

and from here on iam trying to do this but i dont know if iam right or wrong

([OH] - .08943M) - .08943 / ([OH]-.08943)*100 = .001789%