STRONG BASE + WEAK ACID
When equal moles of sodium hydroxide, NaOH, and acetic acid CH3COOH, were mixed they would also react completely
NaOH + CH3COOH � Na+ + {CH3COO}‾ + H2O
However {CH3COO}‾ can itself participate in an equilibrium reaction with water, behaving like a base to accept a proton from water.
{CH3COO}‾ + H2O � CH3COOH + OH‾
It is known as the conjugate (or corresponding) base of acetic acid. So there will be a little acetic acid left no matter what. It is not possible to achieve a stoichiometric point. The final pH of this solution would be slightly basic (highly than 7). For such a titration it is best to use an indicator which changes colour at a pH higher than 7. Like phenolphthalein.
It is also not possible to achieve a stoichiometric point. This solution would be slightly acidic (pH less than 7). The best indicator to use would be methyl red.
WATER
Water has both acidic and basic properties; or amphiprotic.
H2O + H2O � {H3O}+ + {OH}‾
The concentration of {H3O}+ in pure water was found to be 1.004x10‾7 moles per litre. So the concentration of {OH}‾ should also be the same. That is water when pure is neutral in terms of acidity. The pH for this solution would then be;
pH = - log10(1.004x10-7) = 7
So when the pH of an aqueous solution is 7 the solution is said to be a neutral solution (in terms of acidity). pH less than 7 is considered acidic and greater than 7 alkaline (or basic).
BUFFERS
Henderson-Hasselbatch Equation
Based on the equilibrium constant of the Bronsted equilibrium, the expression for pH would be,
pH = pKa - log10{[HB]/[B‾ ]} ; where pKa = - log10{Ka}
This is known as the Henderson-Hasselbatch Equation.
Let us see what happens in a solution containing a weak acid (HA) and the salt of its conjugate base (M+A‾). We will choose a MA that dissociate fully in water (meaning very soluble salts). Let us prepare the solution such that [HA]0 = [M+A‾]0.
The subscribe "0" denote the value at time=0, or initial value. Such values are always known value. Once the two are mixed, two equilibriums would come into play
HA + H2O � {H3O}+ + A ‾
M+A ‾ � M+ + A ‾
HA is a weak acid so the equilibrium is very much to the left and for M+A‾ it will be very much to the right. The concentration of A‾ would push the equilibrium for the dissociation of HA even further to the left. So we can assume that at anytime the concentration of HA is [HA]0 and the concentration of A‾ is [MA]0. Using this value for the Henderson-Hasselbatch Equation we will get
pH = pKa(HA) - log {[HA]0 / [MA]0} = pKa
So we can prepare a solution with known pH using this technology. The best is yet to come.
Let us take the case of phosphoric acid / potassium biphosphate; H3PO4 / KH2PO4, where [H3PO4]0 = [KH2PO4]0 = 0.200M. The pKa value for phosphoric acid is 2.12. So the pH of the solution would be 2.12.
Suppose we accidentally drop some HCl so that its concentration in the mixture is 0.001M. The equilibriums that come into play would be;
HCl + H2O � H3O+ + Cl‾
H3O+ + A‾ � HA + H2O
Consequently [HA] = (0.200 + 0.001) M and [MA] = (0.200 - 0.001) M, the Henderson-Hasselbatch Equation will then be;
pH = pKa(HA) - log { 0.201 / 0.199 } = pKa - 0.004 = 2.12
The pH hardly changes. For a 0.001M HCl solution the pH should have been 3. Such phosphoric acid / potassium biphosphate solution is known as a buffer.
We can prepare various buffers (for various pH value) by using the combination of a weak acid and the salt of its conjugate base (or weak base and the salt of its conjugate acid).
pH buffers are widely used in industry if the reaction is pH sensitive. One good example will be the detergent industry. Most of the detergents have built in pH buffers to keep the pH at a specific value. The next time you see a facial cleansing soap pay attention to this.