Category: Chemical Engineering Math
"Published in Newark, California, USA"
To determine the Ka of a weak acid, a student took 20 mL of a solution of the acid and titrated it with a base. He found out that it took 32 mL of base to neutralize the acid. He then took a new 20 mL of acid and added exactly 10 mL of the same base to it. He measured the pH of this solution and found it to be 4.20. What is the value of Ka for the weak acid?
Solution:
In the given problem, the concentrations of a weak acid and a base are not given but their volumes are given. We can still solve for the value of Ka for the weak acid. Let's assume that the unknown weak acid is a monoprotic acid and the unknown base is a monoprotic base so that it will be easy for us to solve for the value of Ka.
After the titration of 20 mL of a weak acid with 32 mL of a base, the amount of weak acid at the start of the reaction is
or
where CBase is the concentration or molarity of a base in millimoles of solute per milliliter of solution. Let's assign HA for the weak acid and BOH for the base.
The equilibrium reaction for the ionization of a weak acid is
and the ionization constant is
At equilibrium,
where x is the mmoles of HA that partially ionized. Hence, the above equation becomes
If 10 mL of the same base is added to a new 20 mL of a weak acid, then the amount of weak acid that is converted into salt BA is
From the reaction of a weak acid with the base, salt BA is formed
By using mole to mole relationship of the reactants and products, if
then it follows that
As mentioned earlier that 10 mL of the base is added to a new 20 mL of a weak base. At equilibrium, the amount of weak acid will decrease and the amount of product which is the A ion will increase. The ionization constant for the weak acid becomes
Let's assume that the value of Ka is less than 1 x 10-3 so that we can neglect x at the numerator and denominator as follows
If the pH of the solution is 4.20, then the value of x which is the concentration of [H+] is
Take the inverse logarithm on both sides of the equation, we have
or
Therefore, the value of Ka for the weak acid is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
Calculate
the pH of a liter of solution that contains 0.1 mole of acetic acid and
0.1 mole of sodium acetate. If few drops of concentrated hydrochloric acid is added to the
solution that would make the solution 0.02 M in HCl if the buffer was
not present, what is the new pH of the solution?
Solution:
A
buffer solution is a solution of weak acid and its salt or a solution
of weak base and its salt. The purpose of making a buffer solution is to
prevent the rapid change of pH if more acid or base is added to the
solution.
Since acetic acid is a weak acid, then it is partially ionized into acetate ion and hydrogen ion as follows
In a liter of solution at equilibrium,
where x is moles of HC2H3O2 that partially ionized. The ionization constant for HC2H3O2 is given by the equation
From Table of Ionization Constants of Acids and Bases, Ka = 1.8 x 10-5 for acetic acid. The amount of HC2H3O2 that partially ionized is
If
0.1 moles of sodium acetate which is completely ionized into 0.1 moles
of sodium ion and 0.1 moles of acetate ion is added to the solution,
then the above equation becomes
The acetate ion from sodium acetate will be added to the acetate ion from acetic acid. Since the value of Ka is less than 1 x 10-3, then we can neglect x at the numerator and denominator as follows
Hence, the concentration of hydrogen ion is
Therefore, the pH of a buffer solution is
If few drops of concentrated hydrochloric acid is added to buffer solution, then acetate ion will react with hydrogen ion from hydrochloric acid to form acetic acid. The amount of acetic acid will increase while the amount of acetate ion will decrease. From the equation of buffer solution
if
0.02 M HCl solution or 0.02 moles of concentrated HCl is added and
assuming that the increase in volume of solution is negligible, then the
above equation becomes
Since the value of Ka is less than 1 x 10-3, then we can neglect x at the numerator and denominator as follows
Hence, the concentration of hydrogen ion is
Therefore, the pH of final solution is
As
you can see that there's a slight change of pH of buffer solution after
the addition of few drops of concentrated hydrochloric acid. Without buffer solution, the
pH of 0.02 M HCl solution is 1.70.
