Category: Chemical Engineering Math
"Published in Newark, California, USA"
"Hard" water contains small amounts of the salt calcium bicarbonate [Ca(HCO3)2] and calcium sulfate [CaSO4, molecular weight = 136 grams/mole]. These react with soap before it has a chance to lather, which is responsible for its cleansing ability. Ca(HCO3)2 is removed by boiling to form insoluble CaCO3. CaSO4 is removed by reaction with washing soda [Na2CO3, molecular weight = 106 grams/mole] according to the following equation:
If the rivers surrounding New York City have a CaSO4 concentration of 1.8 x 10-3 grams/liter, how much Na2CO3 is required to "soften" [remove CaSO4] the water consumed by the city in one day [about 6.8 x 109 liters]?
Solution:
The first thing that we have to do is to get the amount of CaSO4 in the rivers surrounding New York City as follows
Moles of CaSO4:
From the given reaction above
Moles of Na2CO3:
Therefore, the amount of Na2CO3 required to "soften" or remove CaSO4 in the rivers surrounding New York City is
In metric tons, the weight of Na2CO3 is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
A chemist dissolves BaSO4 in pure water at 25°C. If Ksp = 1 x 10-10, what is the solubility of barium sulfate in water?
Solution:
Since barium sulfate is a slightly soluble salt in water, then it is not 100% completely dissolved. Usually, if a salt is slightly soluble in water, then the solution is cloudy at most. There's also a precipitate of undissolved salt at the bottom of the solution. Ksp will tell us how soluble the salt is. If Ksp = 0, then the salt is not soluble in water at all at any temperature. If you will shake the solution, it is cloudy at first and then later all particles will precipitate at the bottom of the solution. There are some salts that are more soluble in warmer solution rather than in room temperature solution. If Ksp = ∞, then the salt is 100% completely dissolved. The solution is clear at all at any temperature.
Let's consider the barium sulfate that is slightly soluble in water
The solubility of barium sulfate is calculated by its equation
The concentration of barium ion and sulfate ion are expressed in molarity.
Let x = be the concentration of barium ion in the solution.
x = be the concentration of sulfate ion also in the solution.
Therefore, the solubility of barium sulfate in water is
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.
