Category: Chemical Engineering Math
"Published in Newark, California, USA"
The ionization constant for acetic acid is 1.8 x 10-5.
(a) Calculate the concentration of H+ ions in a 0.10 molar solution of acetic acid.
(b) Calculate the concentration of H+ ions in a 0.10 molar solution of acetic acid in which the concentration of acetate ions has been increased to 1.0 molar by addition of sodium acetate.
Solution:
Since acetic acid is a weak acid, then the ionization is not 100% complete. Some molecules of acetic acid will be ionized into acetate and hydrogen ions and the rest of the molecules of acetic acid will remain the same at the end of ionization process. The ionization process of acetic acid is
and its equation for getting the ionization constant is
The concentration of hydrogen ion, acetate ion, and acetic acid are expressed in molarity.
(a) At the start of the ionization process
Let 0 = be the concentration of [H+]
0 = be the concentration of [C2H3O2-]
0.1 = be the concentration of [HC2H3O2]
At the end of the ionization process
Let x = be the concentration of [H+]
x = be the concentration of [C2H3O2-]
0.1 - x = be the concentration of [HC2H3O2]
Therefore, the concentration of [H+] at the end of the ionization process is
Since the value of Ka is very small, then we can neglect x at the denominator so that we don't have to use quadratic formula. If the value of Ka is 1 x 10-3 and above, then we have to solve for the value of x by quadratic formula.
Hence, the value of x which is the concentration of [H+] is
(b) After the addition of sodium acetate to the solution, the concentration of acetate ion has been increased to 1.0 M. Therefore, the concentration of [H+] is
Again, we can neglect x at the denominator since the value of Ka is very small.
Hence, the value of x which is the concentration of [H+] is
Did you noticed that the amount of [H+] becomes smaller after the addition of sodium acetate? Well, if you increase the concentration of either one of the reactants and products, then the concentration of the other substances or components will be affected because of Le Chatelier's Principle.
Category: Chemical Engineering Math
"Published in Newark, California, USA"
The following reaction
was allowed to proceed to equilibrium. The contents of the two-liter reaction vessel were then subjected to analysis and found to contain 1.0 mole H2S, 0.20 mole H2, and 0.80 mole S2. What is the equilibrium constant Keq for this reaction?
Solution:
Since the amount of H2S, H2, and S2 are given already in moles and they are in a 2L reaction vessel, then we can calculate their concentration in molarity.
Molarity of H2S:
Molarity of H2:
Molarity of S2:
Consider the given reaction
The equilibrium constant for the reaction is
The coefficient of each substance in the reaction is equal to the exponent of each factor for the equation above. The units of concentration of substances are always in molarity. Therefore, the equilibrium constant for the reaction is
Note: Usually, the value of Keq is unitless because the units of the concentration of substances are not included in the substitution to the equation of chemical equilibrium. If Keq = 0, then no reaction nor conversion of reactants to products involved in the process and on the other hand, if Keq = ∞, then the reaction is 100% complete in the process.
Category: Chemical Engineering Math
"Published in Newark, California, USA"
What is the maximum weight of SO3 that could be made from 25.0 grams of SO2 and 6.0 grams of O2 by the following reaction? (Atomic Weights: S = 32 and O = 16)
Solution:
The first thing that we need to do is to get the molecular weights of SO2, O2, and SO3 from the given atomic weights.
Molecular Weight of SO2:
Molecular Weight of O2:
Molecular Weight of SO3:
Next, we will calculate the amount of SO3 which is the product of SO2 and O2. From the given reaction
Weight of SO3 from SO2:
Weight of SO3 from O2:
From the two reactants, O2 will give us the least amount of product which is SO3. O2 is the limiting reagent. On the other hand, SO2 is the excess reagent. Once O2 is completely converted into SO3, there will be a left over or unused amount of SO2 in the reaction.
In this problem, we have to choose the limiting reagent to determine the amount of product produced. Therefore, the maximum weight of SO3 is
