Hydrolysis and Buffers
Purpose: Test pH of solutions and then calculate initial concentrations of species involved.HYDROLYSIS
If the salt of a weak acid and a strong base, such as sodium acetate, is dissolved in water, an acid-base reaction involving the anion of the weak acid will occur to generate the acid and hydroxide ion. This type of reaction is referred to as hydrolysis.
H2O + CH3COO- <--> CH3COOH + OH-
As in any other acid-base reaction the extent to which this reaction will occur is determined by the relative strength of the two bases (or the two acids). Since OH- is a strong base (H2O is a very weak acid), hydrolysis does not proceed far to the right unless the salt of a very weak acid is used. Nevertheless, a solution of sodium acetate is slightly basic. Hydrolysis can also occur when the salt of a strong acid and a weak base such as ammonium chloride is dissolved in water. The resulting solution is acidic.
NH4+ + H2O <--> H3O+ + NH3The equation for the hydrolysis of the ammonium ion is
The equilibrium constant expression for hydrolysis of the ammonium ion is equal to Kw/Kb and the magnitude of Ka depends upon the relative strengths of H3O+ and NH4+ as acids.
Hydrolysis can also occur with polyvalent metal ions to generate acidic solutions. Cations in aqueous solutions are bonded to water molecules forming complex ions such as Al(H2O)63+. The higher the charge on these ions, the greater the tendency to hydrolyze to form acidic solutions.
Al(H2O)63+ + H2O <--> H3O+ + Al(H2O)5OH2+BUFFERS
A mixture of relatively large amounts of weak acid and its salt or a weak base and its salt is known as a buffer. Buffers have the property of maintaining a relatively constant pH even when considerable acid or base has been added.
For example, a solution which contained 1.00 M acetic acid and 1.00 M sodium acetate would be a buffer whose H+ concentration would be 1.8 x 10-5 M and have a pH of 4.74.
Ka = [CH3COO-][H+]/[CH3COOH] = 1.8 x 10-5
If 0.05 mole of HCl were added per liter of solution, the CH3COOH concentration would become 1.05 M and the CH3COO- concentration would become 0.95 M. The new H+ concentration can be calculated using the Ka expression:
The pH of this solution is 4.70.
In a similar manner, if 0.05 mole of NaOH were added to the original solution, the H+ concentration would be:
and the pH would be 4.78. In other words, the pH would
only change by .04 in the buffer solution in each case. This
relative pH stability would be true until one had exceeded
the capacity of the buffer to absorb H+ or
For comparison, if 0.05 mole of HCl were added to one liter of pure water, the [H+] of the resulting solution would be 5.0 x 10-1 M and the pH would be 1.3. Thus, for the buffered solution the pH changed by .04 units whereas for water it changed by 5.7 units.Procedure:
Lab report:Very simple. For each section, give short easy to understand statements about what you did, then give the appropriate calcualtions. You need to include things such as how much of each thing you combined. You are walking me through what you did in lab, in the order you did it. Make sure it is understandable, use complete sentences.
Prelab Questions: (worth 8 points)
1. Calculate the pH of a 0.87 M NaCN solution.
HINT: NaCN dissociates into Na+ (an unexciting acid) and CN-, the conjugate base of HCN. You can look up the Ka for HCN, and therefore calcuate the Kb for the BASE, CN-. It is this equation (Kb) that is your equilbirum expression for this problem.
2. What is initial concentration of a sodium acetate if the pH (at equilibrium) is measured to be 9.30? HINT: Same sort of thing as question 1, sodium acetate dissociates into Na+ and acetate ion (a base)......
~MEO 22 Mar 06