What is the change in ph after addition of 10.0 ml of 1.0 m sodium hydroxide to 90.0 ml of a 1.0 m nh3/1.0 m nh4+ buffer? [kb for ammonia is 1.8 x 10-5]?

Answer 1

I assume you mean M(molar) and not m(molal).
First, calculate the pH of the solution before any NaOH is added. Use the Henderson-Hasselbalch equation.
pH = pKa + log (NH3)/(NH4^+)

Then calculate the pH after the addition of the NaOH. I would use millimols for this because it makes it easier than using concentration. Technically, you're supposed to use concentration but the volume of the solution cancels so the answer using mmols and the answer using concentration is the same.
mmols NH3 = mL x M = 90
mmols NH4^+ = mL x M = 90
mmols NaOH added = mL x M = 10
.....NH4^+ + NaOH ==> NH3 + H2O + Na^+
I....90.......0.......90..........
add...........10...............
C...-10......-10......+10
E....80.......0........100

Plug the E line into the HH equation and solve for pH.
Subtract the two pH values to find the difference.