A buffer is a solution that resists changes in pH when small amounts of H+ or OH- are added. This system acts as a buffer because the ammonia reacts with acid and the ammonium ion reacts with base:
NH3 + H+ -> NH4+
NH4+ + OH- -> NH3
The pH of a buffer can be calculated using the Henderson-Hasselbalch equation. Here's the form of the equation for a base and its conjugate acid:
pOH = pKb + log [BH+]/[B], pH = 14.0 - pOH
In this case the base B is NH3 and the conjugate acid BH+ is NH4+.
You need to know the Kb for ammonia, which is 1.8 x 10^(-5).
The initial pH of the buffer is:
pOH = -log[1.8x10^(-5)] + log(0.36)/(0.30)
pOH = 4.74 + 0.079 =4.819, pH = 14.0-4.189 = 9.18
The NaOH that's added reacts the NH4+, converting it to NH3.
We can find the new concentration of NH4+ by subtracting the amount that reacted:
moles NH4+ reacted=moles OH- added =(0.050M)(0.0200 L)=(0.00100 moles)
moles of NH4+ originally in 80.0 ml = (0.36M)(0.0800 L) = 0.0288 moles
moles NH4+ left = 0.0288-0.00100 = 0.0278 moles
new [NH4+] = 0.0278 moles/(0.0200+0.0800)L = 0.278 M
Now we will find the new concentration of NH3 by adding the amount formed:
moles NH3 formed = moles OH- added = 0.00100 moles
moles of NH3 originally in 80.0 ml = (0.30M)(0.0800 L)=0.024 moles
moles of NH3 after addition of OH- = 0.024+0.00100=0.025 moles
new [NH3] = 0.025 moles/(0.0200+0.0800)L = 0.250M
Now plug the new concentrations into the Henderson-Hasselbalch equation:
pOH = 4.74 + log(0.278/0.250)
pOH = 4.74 + 0.046 = 4.786, pH = 14.0-4.786 = 9.21