Home > Chemistry > Chemistry 1A Part Deux: Lecture 17 Notes

Chemistry 1A Part Deux: Lecture 17 Notes


Today is a 51 minute lecture on a very special case of acid base chemistry, buffers.  Other listener has complained that this material complex – it is!  To become proficient, you must write reactions and work problems.  Listening is not quite enough here.

Solubility equilibria is up next.


Buffers resist changes in pH upon addition of strong acids/bases.  This is important in oxygen transport in the body.  Arteries carry oxygenated blood, veins carry deoxygenated blood.

The known can be in the flask.  It is preferable to put it in the buret.

CO2 + H2O = HCO3- + H+

HbH+ + O2 = HbO2 + H+.

When you don’t have enough oxygen, you make lactic acid.  This occurs during blood loss, leading to acidosis or shock.  The patient begins to pant as they try to get rid of CO2.

To calculate the acid concentration,  we can use the equivalence point to find the initial concentration by drawing a line straight down and noting where the x axis is intercepted.

Moles base  = final volume x concentration = a number…

Moles acid = concentration/ initial volume = another number

We can also do it from the initial pH.

pH = -log [H+]

[H+] = 10^-pH

Strong acids are easy.  For weak acids, you’ll have to do equilibrium, knowing Ka.

Ka = [A- H+/ HA

Different acids with the same concentration.  Halfway to the equivalence point is a very special place.  There pH = pKa.  At the equivalence point itself, the pH is The weaker the acid, the stronger the conjugate base.

The buffer region doesn’t change much for weak acids before the equivalence point.

HA + OH- = H2O + A-

Addition of base removes H+ and converts HA into A-.  Alot of HA remains.

HA = H+ + A-

Ka = [H+] [A-]/[ HA]

Ka = H+ when A- = HA

pH = pKa + log (base/acid)


Buffers contain large amounts of both weak acids and its conjugate base.  They resist large changes of PH upon addition of strong acids/bases.

HA = H+ + A-

pH = pK + log (base/acid)

             HA =         H+      +       A-

I           Initial          0                 N

C            -x               x                 x

E           I – x             x          N +x

For dry ice in a basic solution:

CO2 + H2O = H+ + HCO3-         Ka = 4.5 x 10 -7

PO4 -3 + H2O = HPO4-2  + OH-        Kb = 2.1 x 10^-2

Add dry ice to NH3 at pH 11.  How does the pH change compared to NaOH at pH 11?

The NH3 requires more ice to neutralize because it has a weaker base.

Making a Buffer Solution

Formic acid is CH3COOH with Ka = 2x 1-5 pKa 4.74

Mix 500 mL of CH3COOH with 500 ml of .5M

 CH3COONa + CH3C00H = CH3COO- + H+

Mix 500 mL of 1M formic acid with 500 mL .5M NaOH

CH3COOH + OH- = CH3COO- +  H2O K >>1

CH3C00H = CH3COO- + H+    Ka

Mix 500 mL 1M CH3COONa with 500 mL .5 M HCl

CH3COO- + H+ = CH3OOH        for K>>1

CH3COO- + H2O = CHCOOH + OH-        Kb


Yagi Antenna: reflector, driver, director, unidirectional


Tissue: epithelial, connective, muscular, nervous

Electronegativity: NOF

PV = nRT

K = products/reactants

G = H – TS

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