The reaction we will look at is :
Al_{(s)} + Fe_{2}O_{3}_{(s)} Al_{2}O_{3}_{(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.
Weak acid dissociations involve equilibria. The equilibrium constants have a specific symbol = K_{a}.
Example: What is the pH of a 0.10 M solution of acetic acid. K_{a} = 1.8 x 10^{5}
K_{a} = [H^{+}][OAc^{}] / [HOAc]
HOAc  H^{+}  +  OAc^{}  
Before reaction  0.10 M  0  0  
@ Equilibrium  0.10 M x  x  x 
assume x << 0.1 since K_{a} =1.8 x 10^{5}, then [HOAc] = 0.10 M
Substituting, K_{a} = (x)(x) / 0.10 = 1.8 x 10^{5},
x^{2} = 1.8 x 10^{6}
x = 1.34 x 10^{3}M; assumption OK.
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.
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. K_{a} = 1.8 x 10^{5}
HOAc  H^{+}  +  OAc^{}  
Before reaction  0.0125 moles/L  0  0.0250 moles/L  
@ Equilibrium 

x 

pH = 5.046
Let's look at a second way to solve buffer problems, using something called the HendersonHasselbalch equation. The HendersonHasselbalch equation is simply a log version of the equilibrium expression for acid dissociation.
K_{a} = [H^{+}][A^{}] / [HA], where HA is the acid and A^{} is its salt (conjugate base)
K_{a} = [H^{+}]([A^{}] / [HA])
If we take logs of both sides of the equation we get:
so
HendersonHasselbalch equation: pH = pK_{a} + log([A^{}] / [HA]) (Note that an analogous equation may be written for bases: pOH = pK_{b} + log([B^{+}] / [BH]))
So now let's do the example from above using the HendersonHasselbalch 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. K_{a} = 1.8 x 10^{5} (pK_{a} = 4.74).
pH = pK_{a} + 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
Syllabus / Schedule 
© R A Paselk
Last modified 3 May 2013