Mole Fraction Problems, 2

Category: Chemical Engineering Math

"Published in Newark, California, USA"

The vapor pressures of pure benzene and toluene at 60°C are 385 and 135 Torr, respectively. Calculate (a) the partial pressures of benzene and toluene, (b) the total vapor pressure of the solution, and (c) the mole fraction of toluene in the vapor above a solution with 0.60 mole fraction toluene.

Solution:

In the given problem, if mole fraction of toluene in a solution is

then mole fraction of benzene in a solution is

The sum of the mole fractions of the components in a solution must be equal to 1. In this case, benzene and toluene are the components in a solution.

The vapor pressures of pure benzene and toluene at 60°C are

(a) By Raoult's Law, we can calculate the partial pressures of benzene and toluene in the vapor. 

Partial Pressure of Benzene:

Partial Pressure of Toluene:

  

(b) The total vapor pressure of the solution is
 

 

  

(c) Since we know already the partial pressures of benzene and toluene as well as the total vapor pressure of the solution, therefore, the mole fraction of toluene in the vapor is
 

 

 

 

Stoichiometry Problem - Combustion

Category: Chemical Engineering Math

"Published in Newark, California, USA"

Four liters of octane gasoline weigh 3.19 kg. Calculate the volume of air required for its complete combustion at STP.

Solution:

The first thing that we have to do is to write the chemical equation for the combustion of octane gasoline as follows

Make sure that the above chemical equation is balanced. Next, we will calculate the molecular weight of octane gasoline as follows

Moles of octane gasoline:

From the chemical equation above, moles of oxygen is

At STP (Standard Temperature and Pressure), 1 mole of any gas is equal to 22.4 liters. Hence, the volume of oxygen is

Since air contains 21% oxygen and 79% nitrogen, therefore, the volume of air required for the complete combustion of octane gasoline is

Stoichiometry Problem - Total Hardness

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

Solubility Product Constant Problems

Category: Chemical Engineering Math

"Published in Newark, California, USA"

A chemist dissolves BaSO₄ in pure water at 25°C. If K_sp = 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. K_sp will tell us how soluble the salt is. If K_sp = 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 K_sp = ∞, 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
 

 

 

 

 

Ionization Constant for Acid Problems

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 K_a 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 K_a 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 K_a 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.