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

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ngruby82

30 de outubro de 2021 08:06

There is a buffer that contains 15mL of 0.10M HOAc and 5mL of 0.10M NaOAc. What is the pH after 1 drop 0.10M NaOH and the pH after 0.10M NaOH? I am trying to do that by first finding the moles of OH- added then moles HAc initial & Moles Ac- initial. Then finding moles of HAc after and moles of Ac- after by doing initial +/- moles OH- added but I am having trouble finding moles of OH- added. Please help.

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Galaxy027❤

25 de novembro de 2021 00:40

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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.)

NaOAc(s) => Na^+(aq) +
OAc^-

(use the reaction equillibrum eqn)

HOAc(aq) + H₂O(l) H₃O^+(aq) + OAc^-(aq)
K_a = 1.7 x 10^-5

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.

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

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