Archive for October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
A laboratory technician wants to determine the aspirin content of a headache pill by acid-base titration. Aspirin has a Ka of 3.0 x 10-4 M. If the pill is dissolved in water to give a solution about 0.005 M, what is the pH of this solution? (Neglect dilution effects.)
If the solution in the problem above is then titrated against KOH solution, what will be the pH at the stoichiometric point.
Chemistry Tutor Site - Chemistry Homework Answer
With this Ka and concentration, you can’t use the usual formula for weak-acid equilibrium
HA -> H+ + A-
HA dissociated into H+ and A-, if the X mol of the HA is dissociated thus the H+ and A- would be X mol too. Thus
X2 / [Acid conc] = Ka
because you can’t neglect the amount of acid actually reacted to H+ and the Cation. So you have the more general equation
X2 /[0.005-X]= 3×10-4
You can solve this by trial/error or by the general quad. eqtn method. Roughly, X=0.0027
At the stoichiometric point, all acid is dissociated, and the problem is equivalent to dissolving 0.005 M potassium salicylate in water. In that case,
Salicylate- + H2O = Acid + OH-
In this situation, the equilibrium expression is
[Acid][OH-]/[Salicylate] = Kb
Since KbKa=10×10-15, Kb= 3.3×10-11/p>
If X is the amount of [OH-] formed at equil., then
X2/[0.005]= 3.3×10-11
X is roughly 4×10-7. From this you can compute pOH= 7-log 4= 6.4. Then since pH+pOH=14, pH=7.6.
October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
Why does the pH drop rapidly when an acid-base titration reaches equivalence point? Why doesn’t the titration curve look more gradual? The acid is being added drop by drop to the base, so shouldn’t the pH fall drop gradually? Why does the PH change so sharply when it reaches the equation ratios? What is going on at an aqueous level?
Chemistry Tutor Site - Chemistry Homework Answer
The weaker the acid or base is the more gradual the titration curve will look. You can see this if you compare the titration curve of acetic acid to that for hydrochloric acid. Acetic acid has a much more gradual curve.
When you are at the equivalence point the pH is about 7 (depending on the acid/base), which means the hydronium ion concentration is 10^-7. If you add an acid or base it will move away from that concentration by orders of magnitude, which causes the sharp up or down (depending what is being tritrated). After a few drops have been added in the titration the curve levels out because you are only doubling or adding to the ion concentration at the same magnitude instead of changing the order of magnitude.
In other words, after the equivalence point is reached, when you add a drop the pH would change from 10^-7 to 10^-6 (if titrating with an acid) or 10^-8 (if titrating with a base. This is a factor of 10 change, and the curve is steep. But later on if you add another few drops you only change the concentration by doubling the ion concentration. So titrating with an acid goes from, say, 1×10^-5 to 2×10^-5 or something similar if titrating with a base.
October 17th, 2008
For a general chemical reaction
aA + bB <-> cC + dD
the equilibrium constant can be defined by
K = [C]c[D]d / [A]a[A]a
It is conventional to put the activities of the products in the numerator and those of the reactants in the denominator. Stability constants, formation constants, binding constants, association constants and dissociation constants are all types of equilibrium constant.
In a chemical process, chemical equilibrium is the state in which the chemical activities or concentrations of the reactants and products have no net change over time. Usually, this would be the state that results when the forward chemical process proceeds at the same rate as their reverse reaction. The reaction rates of the forward and reverse reactions are generally not zero but, being equal, there are no net changes in any of the reactant or product concentrations. This process is called dynamic equilibrium
October 17th, 2008
Chemistry Tutor Site - Chemistry Homework Help
What is the equilibrium concentration of HI in the gaseous mixture?,/p>
H2(g) + I2(g) ↔ 2HI Temperature = 731K
1.40 mol of H2 and 1.40 mol of I2 are placed in a 1.00 L vessel. What is the equilibrium concentration of HI in the gaseous mixture?
The equilibrium constant is K = 49.0
Chemistry Tutor Site - Chemistry Homework Help
The reaction:
H2 (g) + I2 (g) <——-> 2HI
initial concentration
1.40 .. . .1.40
at equilibrium
1.40-x. .. . 1.40-x .. . . ..2x
49.0 = (2x)2 / (1.40-x)sup>2
Square root
7 = 2x / 1.40-x
x = 1.09 M
2x = [HI] = 2.18 M
October 17th, 2008
Previous Posts