Finding pH of 1 drop and 10 drops of 0.10M NaOH

Hello ngruby82!

So firstly, you can calculate the pH of a buffer using the acid dissociation constant expression:

Therefore, NaOH (NaOH + H2O)  = Na+ and OH-ions. So due to NaOH being a strong base, it ionizes completely in water. 

Buffer solution = dn/d(pH)

Therefore, 0.10 mol of NaOH will separate into 0.10 mol of Na+ and OH-.

(For the pH calculation formula) 

pOH = -log[OH-]

-log[0.10] = 1.

(We know the pH scale = 14)

Therefore 14 - 1 = 13

therefore your pH of NaOH = 13.

So if 0.10 M HOAc:[H₃O^+] = 1.3 x 10^-3

MpH = 2.8

0.10 M NaOAc:[OH^-] = 7.5 x 10^-6

MpH = 8.9

(give the H₃O^+ and OAc^- ions.)

HOAc + H₂O <=> H₃O^+ (+ OAc^-)

(NaOAC breaks down in water so give the OAc^- ion.)

(use the reaction equillibrum eqn)

delta-C  = 1.7 x 10^-5

(Therefore the solution is acidic but less than HOAc) =

pH = - log [1.7 x 10^-5] = 4.56

- Hope this helps!

~Galaxy.