pH
The pH = - log [H3O]+ is a measure of the hydronium concentration of an aqueous solution. The lower the pH the more acidic the solution. A table summarizing log values for various powers of ten is shown below.
| # | 1000 | 100 | 10 | 0.1 | 0.01 | 0.001 |
| scien.not. | 103 | 102 | 101 | 10-1 | 10-2 | 10-3 |
| log | 3 | 2 | 1 | -1 | -2 | -3 |
| -log | -3 | -2 | -1 | 1 | 2 | 3 |
A 0.1 M HCl solution has 0.1 M H3O+ and a pH = 1, whereas a 0.1 M acetic acid solution (CH3COOH) has 0.001 M H3O+ (since it is only 1% dissociated) and a pH = 3
The Keq of an acid is a measure of the strength of an acid. For the general acid reaction with water:
HA(aq) + H2O(l) <==> H3O+ + A-, Keq = ([H3O+][A-])/([HA][H2O])
For a strong acid, Keq > 1 and for a weak acid, Keq < 1.
[H2O] is approximately 55.5 M in a diliute acid
solution, and is essentially constant (since it is present in such great excess). Hence Keq[H2O]
= ([H3O+][A-])/[HA] = Ka.
Ka, the acid constant, is also a constant; for a strong acid, Ka > 1 and for a weak acid, Ka < 1
Consider the two reactions below in which the concentration of HCl and CH3COOH are each 0.1 M.:
HCl + H2O ----------> H3O+ + Cl-
At equilibrium, [H3O+] and [Cl- ]
are both approximately 0.1 M and HCl is effectively 0. In actuality, some HCl is left and
can be shown to be 0.000000001 M = 10-9 M.
Hence Ka = ([H3O+][A-])/[HA] = (10-1)(10-1)/10-9
= 107 >> 1.
CH3COOH + H2O ----------> H3O+ + CH3COO-
At equilibrium, [H3O+] and [Cl- ]
are both approximately 0.001 M and CH3COOH is 0.099 M = 9.9 x10-2 M.
Hence Ka = ([H3O+][A-])/[HA] = (10-3)(10-3)/9.9
x 10-2 = 1.01 x 10-5 << 1.
Just as pH = -log[H3O]+, pKa = -logKa. In the above examples, the pKa of HCl is -7 while that of acetic acid is approx. 5.
The pKa of an acid is a measure of the strength of an acid. It is a constant and does not depend on the concentration of the acid. The lower the pKa, the more stronger the acid.
Water can act as both a weak acid and a weak base. It can interact with itself to form the hydronium and hydroxide ion, as shown below:
H2O + H2O ----------> H3O+ + OH-
In pure water, [H3O+] and [OH-] are equal (as seen in the equation above) and are each 1x10-7 M. Hence the pH of neutral water is 7.0
Buffers
Buffer solutions consist of weak acids and salts of weak acids. They can be created by mixing a solution of a weak acid (such as acetic acid) with a solution of a salt of a weak acid (such as sodium acetate). Alternatively, buffer solutions can be created by partially titrating a solution of a weak acid with sodium hydroxide which neutralizes part of the acid to form the salt of the weak acid, with the sodium ion provided by the added titrant.
Buffer solutions can neutralize the addition of small amounts of an acid (which reacts with the base product of the weak acid) and small amounts of a base (which reacts with the weak acid).
Consider what would happen to the pH of pure water (pH = 7) if either 0.01 mol of NaOH or 0.01 mol of HCl were added to sufficient water to make 1L. The first solution would now have an [OH-] = 0.01 M and hence a [H3O+] = 1 x 10-12 M, or a pH = 12. The second solution would have a [H3O+] = 1 x 10-2 M, or a pH = 2. This examples shows that the addition of a small amount fo either acid or base to water dramatically changes the pH. Consider the pH of pure water to be like a balance. Addition of small amounts of either acid or bases dramatically changes the balance (pH)
Animation: Addition of NaOH and HCl to water
Now consider what would happen to the pH of a 0.5 M acetic acid (a weak acid) solution but which also contains 0.5 M sodium acetate (a salt of a weak acid). (Recipe: 5.8 ml glacial AA + 17.6 g NaAc.3H20 to 200 ml total)
The appropriate equation to describe this solution is:
CH3COOH + H2O <==> H3O+ + CH3COO-
The pH of the solution can be calculate if you know the Ka of acetic acid = 1.8 x 10-5. Plug this into the following equation to get x = [H3O+] and pH as follows:
Ka = ([H3O+] [CH3COO-])/[CH3COOH] = x(0.5)/0.5 = 1.8 x 10-5
x = [H3O+] = 1.8 x 10-5; pH = 4.74
Now if 0.01 mol of NaOH was added (without changing the volume), the hydroxide would react with the weak acid as shown below:
CH3COOH + OH- <==> H2O + CH3COO-
Therefore the base is consumed, CH3COOH is reduced to 0.49 M and CH3COO- increases to 0.51. As above, x can be calculated and the pH determined.
Ka = ([H3O+] [CH3COO-])/[CH3COOH] = x(0.51)/0.49 = 1.8 x 10-5
x = [H3O+] = 1.72 x 10-5; pH = 4.76.
The pH was raised only by 0.02 units in comparison to the case of pure water whose pH increased by 5 units.
Now if 0.01 mol of HCl was added (without changing the volume), the HCl would react with the weak base acetate as shown below:
HCl + CH3COO- <==> Cl- + CH3COOH
Therefore the added HCl is consumed, CH3COO- is reduced to 0.49 M and CH3COOH increases to 0.51. As above, x can be calculated and the pH determined.
Ka = ([H3O+] [CH3COO-])/[CH3COOH] = x(0.49)/0.51 = 1.8 x 10-5
x = [H3O+] = 1.87 x 10-5; pH = 4.73. The pH was reduced only by 0.01 units in comparison to the case of pure water whose pH dropped by 5 units.
Animation: Addition of NaOH and HCl to an acetic acid/sodium acetate buffer
Biological Buffer Systems:
pH in the body has to be very tightly controlled. Specifically, the pH of blood and intracellular spaces can not very much without serious consequences. Two common buffers systems are
The later is most important in blood, which is maintained within a narrow range of 7.35-7.45. This buffer system is a bit more complicated since another reaction of carbonic acid must be considered - namely the breakdown of carbonic acid to water and CO2.
CO2(g) + H2O(l) <==> H2CO3(aq) + H2O(l) <==> H3O+(aq) + HCO3-(aq)
I demonstrated the reversible formation of carbonic acid in class by blowing into a water which was made slightly alkaline by the addition of a small amount of ammonia. The solution become more acidic, as indicated by the change of color of the solution when phenopthalein was added. The reaction of carbon dioxide with water to form carbon dioxide can be visualized below:
The two major components of the buffer system of blood are controlled by two different mechanisms:
H2CO3 is controlled by changing the respiration rate. If you breath more quickly and deeply (like when you hyperventiallate), you exhale more CO2. This will decrease the blood carbonic acid by as shown in blue below. CO2 concentrations decrease, pulling the equilibria below to the left which decreases H3O+(aq)
CO2(g) + H2O(l) <==> H2CO3(aq) + H2O(l) <==> H3O+(aq) + HCO3-(aq)
HCO3-(aq) (bicarbonate) is controlled by the rate at which it is excreted by the urine.
What happens to the blood pH when the relative amounts of H2CO3(aq) and HCO3-(aq) change. Lets examine the Ka for the red part of the equation below, considering carbonic acid as a weak acid.
CO2(g) + H2O(l) <==> H2CO3(aq) + H2O(l) <==> H3O+(aq) + HCO3-(aq)
Ka = ([H3O+][HCO3-])/[H2CO3] which can be rearranged to
[H3O+] = Ka[H2CO3] /[HCO3-]. From this is should be clear that:
Two kinds of acidosis are common - respiratory acidosis and metabolic acidosis.
Metabolic acidosis: This occurs often in burn patients. Blood plasma leaks into damaged areas which can decrease total blood volume and flow, decreasing oxygen availability. This will increase anaerobic metabolism, just as when you are running a fast race and are deprived of oxygen. Lactic acid builds up in the blood, increasing the H3O+ in the blood, shifts the equilibrium shown in red below to increase carbonic acid and decrease bicarbonate:
CO2(g) + H2O(l) <==> H2CO3(aq) + H2O(l) <==> H3O+(aq) + HCO3-(aq)
This decreased pH causes the injured person to break harder to get rid of CO2, which by Le Chateliers principle causes a helpful decrease in carbonic acid as shown in blue below:
CO2(g) + H2O(l) <==> H2CO3(aq) + H2O(l) <==> H3O+(aq) + HCO3-(aq)
If the damage is to great so the person can't breath well, shock can develop.
Respiratory acidosis: This occurs in burn victims who have inhaled much smoke and can't breath well. Likewise it occurs in pulmonary diseases such as emphysema or a physical obstruction which hinders respiration. This causes CO2 to build up, increasing carbonic acid and decreasing blood pH as shown below in red:
CO2(g) + H2O(l) <==> H2CO3(aq) + H2O(l) <==> H3O+(aq) + HCO3-(aq)