Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2013

Lecture Notes 37: 3 May

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Thermite Demonstration

The reaction we will look at is :

Al(s) + Fe2O3(s) right arrow Al2O3(s) + Fe(l)

This reaction has a very high activation energy, the thermite powder is quite stable, so we have to add lots of heat to get it started. To do that today we will use the energy generated by a burning Mg ribbon to initiate the thermite reaction. Once started the reaction itself liberates an immense amount of energy as heat, which allows the reaction to proceed to completion and gives molton iron as the product.

pH of weak acid solutions

Weak acid dissociations involve equilibria. The equilibrium constants have a specific symbol = Ka.

Example: What is the pH of a 0.10 M solution of acetic acid. Ka = 1.8 x 10-5

Ka = [H+][OAc-] / [HOAc]

  HOAc  equilibrium arrow H+ + OAc-
Before reaction 0.10 M   0 0
@ Equilibrium 0.10 M- x   x   x 

assume x << 0.1 since Ka =1.8 x 10-5, then [HOAc] = 0.10 M

Substituting, Ka = (x)(x) / 0.10 = 1.8 x 10-5,

x2 = 1.8 x 10-6

x = 1.34 x 10-3M; assumption OK.

pH = - log (1.34 x 10-3) = 2.87

Notice the significant figures. For a log function the number in front of the decimal is the exponent of ten, thus pH = 2.87 is a 2 significant figure number.

Acid Equilibria

Buffer calculations: One of the most frequent calls for calculating acid equilibria is calculations involving buffers. What is a buffer?

With this in mind let's do some examples.

Example: Calculate the pH of a buffer made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. Ka = 1.8 x 10-5

  HOAc  equilibrium arrow H+ + OAc-
Before reaction 0.0125 moles/L   0 0.0250 moles/L
@ Equilibrium
 
(0.0125 - x) M
assume x is small,
= 0.0125
  x  
 
(0.0250 + x) M
assume x is small,
= 0.0250

Ka = [H+][OAc-] / [HOAc]

Substituting, Ka = [H+](0.0250) / (0.0125) = 1.8 x 10-5

Rearranging, [H+] = (1.8 x 10-5)(0.0125) / (0.0250) = 0.90 x 10-5

x is within experimental error (0.000009 < ±0.0001), so assumption OK

pH = 5.046

Let's look at a second way to solve buffer problems, using something called the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation is simply a log version of the equilibrium expression for acid dissociation.

Ka = [H+][A-] / [HA], where HA is the acid and A- is its salt (conjugate base)

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

If we take logs of both sides of the equation we get:

log Ka = log[H+] + log([A-] / [HA])

Rearranging: - log[H+] = - log Ka + log([A-] / [HA])

But - log[H+] = pH, and by analogy -log Ka = pKa!

Substituting, pH = pKa + log([A-] / [HA])

so

Henderson-Hasselbalch equation: pH = pKa + log([A-] / [HA])

(Note that an analogous equation may be written for bases: pOH = pKb + log([B+] / [BH]))

So now let's do the example from above using the Henderson-Hasselbalch equation.

Calculate the pH of a buffer made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. Ka = 1.8 x 10-5 (pKa = 4.74).

pH = pKa + log[A-] / [HA]

pH = 4.74 + log(0.0250) / (0.0125)

= 4.74 + 0.301

pH = 5.04

Notice that only the ratio matters, the actual concentrations of the acid and salt are not that important for determining pH!

What is the pOH of this solution? Use relationship pH + pOH = 14.00 {from [H+][OH-] = 1.0 x 10-14 , log[H+][OH-] = log(1.0 x 10-14), etc.}

then pOH = 14.00 - 5.04 = 8.96

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© R A Paselk

Last modified 3 May 2013