Adding strong base to basic buffer solution.?

What will be the pH change when 20.0mL of 0.100 M NaOH is added to 80.0mL of a buffer solution consisting of 0.169 M NH3 and .183 M NH4CL?



I calculated that before the NaOH is added the pH is 9.29 but I don't know how to do the rest.Adding strong base to basic buffer solution.?
Okay, so what you have to do here is find the moles of NaOH and the moles of your initial acid and base, find the new molarities after the change, and use the henderson-hasselbalch equation with your new concentrations.



so for the initial pH, i used pKa of NH4Cl (9.3).



pH = 9.3 + log (0.169M/0.183M), and I found the pH to be 9.265, which is a little different.





then, I found the moles of NaOH using M = mol/L



0.1M = x mol/0.02L, and found the moles to be 0.002.



after finding the moles of NaOH, i did the same method to find the moles of NH3 and NH4Cl, which were 0.0135 and 0.01464, respectively.



when adding a strong base to a buffer, the acidic component reacts with the added base, causing OH- ions to be released and the moles of acid to decrease and base to increase, respectively.



with this knowledge, I added the number of moles of NaOH to the moles NH3

0.0135 + 0.002 = 0.0155 moles



and subtracted the number of moles of NaOH from the moles of NH4Cl

0.01464 - 0.002 = 0.01264 moles



i then found the molarities of each by dividing the moles by (0.08 L + 0.02 L to account for the addition of NaOH), and found the new concentrations.



[NH3] = 0.155M, [NH4Cl] = 0.1264M



now that we have the new concentrations of acid and base, substitute them into the H-H equation to get the new pH:



pH = 9.3 + log (0.155/0.1264) = 9.39.



9.39 - 9.265 = a pH change of +0.125.



there may be some minor errors in there, but that is how one generally figures out these problems. good luck!