millimols NH3 initially = mL x M = 50 x 0.18 = 9
millimola HBr initially = 5 x 0.36 = 1.8
..............NH3 + HBr ==> NH4Br
I.............9..........1.8...........0
C.........-1.8......-1.8.........+1.8
E............7.2.........0...........1.8
So your solution has 7.2 mmols NH3 and 1.8 mmols of its salt of NH4B4. That is a buffered solution so apply the Henderson-Hasselbalch equation and calculate pH.
pH = pKa + log (base)/(acid)
Convert Kb for NH3 to Ka then to pKa.
M base = M NH3 = mmols/total mL
M acide = M NH4^+ = mmols/totalmols.
Post your work if you get stuck.
E........
50.0 mL of 0.18 M NH3 with 5.0 mL of 0.36 M HBr.
Kb of NH3 is 1.77 x 10-5
Calculate pH
I don't know where to begin
4 answers
7.2mmols/55ml = .1309 NH3
1.8 mmols/ 55ml = .0327 NH4
-log(1.77 x 10-5) = 4.752
14-4.752 = pKa 9.24798
9.24798 + log (.1309/.0327) = 9.8503
1.8 mmols/ 55ml = .0327 NH4
-log(1.77 x 10-5) = 4.752
14-4.752 = pKa 9.24798
9.24798 + log (.1309/.0327) = 9.8503
is that right^
That looks OK to me but you aren't allowed that many places in your answer. I would round it to 9,85.