1. Short Logarithm review:Since Ihave your undividedattention,let’sgooverlogsquick.
The onlythingyou needtoknowaboutlogsis thisone equation. Youcan figure everythingelse out
fromhere.
Log(10) = 1 or log(101
) = 1
or 10x
= 10; x = 1
HendersonHasselbalch
The shape of the titrationcurve of any weakacidis describedby the Henderson-Hasselbalchequation,
whichisimportantforunderstandingbufferactionandacid0base balance inthe bloodandtissue of
vertebrates.
pH = pKa + log([A-]/[HA])
pKa = -log(Ka),where Ka isthe midpointinthe titrationof the weakacid. Whatis the pH equal to at this
point?
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This equation also allowsusto:
1) Calculate the pKa,givenpHand molar ratioof protondonorand acceptor
2) Calculate pH,givenpKa and the molar ratioof protondonorand acceptor
3) Calculate the molarratioof protondonorand acceptor,giventhe pH and pKa.
A bufferismade from100ml of 1.0M acetic acid(Ka = 1.8 x 10–5
) and50ml of 5.0M the aceticion which
isacetic acid’sconjugate base.Calculate the pHof thisbuffer.
Since the twoequationswe needare givenabove,let’slookatwhatwe needtocalculate. The question
clearlysaysto calculate the pH. Sowe shouldimmediatelybringupthe equationthathaspH in it.
pH = pKa + log([A-]/[HA])
Since we’re solvingforpH,it shouldfollowthatwe can calculate all the othervaluesinthisequation.
Letsfirstcalculate the pKa. givenbythisequation:
pKa = -log(Ka)

2. it’sa simple calculationtodiscoverthe pKa. by plugginginthe givenKa of aceticacidinto the equation
we get:
pKa = -(log(1.8 x 10–5
)) = -(log(1.8)+log(10-5
))
=-(.255+(-5))= -(-4.745) = 4.745
pKa = 4.745
Nowwe needtofindout whatvalueswe can putin for[A-] and[HA]. [HA] refersto the concentration
of the acid. So 1.0M inthe problemisreferringtomolarity. Sowe have 100ml of 1.0M of our acid.
Molarityismols/liter. Soif we multiplythe molaritybythe litersof acidthatwe have,we shouldget
howmany molesof the acidwe are adding. Thiscan be done withthe conjugate base as well.
[HA] [A-]
100 ml = 0.1 L 50 ml = 0.05 L
0.1 L x 1.0 mols/liter= 0.1 mols ofacetic acid 0.05 L x 5.0 mols/liter= 0.25 mols ofacetic
ion
[HA] = 0.1 [A-]= 0.25
Nowthat we have all the valuesthatwe need,it’sa simple mattertothrow themintothe equationand
getan answer.
pH = pKa + log([A-]/[HA])
pH = 4.745 + log(0.25/0.1)
pH = 4.745 + log(2.5)
pH = 4.745 + 0.3979
pH = 5.143
..........
Usingwhat you’ve learned,calculate the pHof a 2.0M of nitrous acid(Ka= 7.2 x 10-4
) and a 2.0M of
aceticionassume that theyhave equivalentvolumes.
pKa = 4.745
[HA] = 0.1
[A-]= 0.25