Solving for pH of a Mixture of Acid and Base Problems, 2

Category: Chemical Engineering Math

"Published in Newark, California, USA"

What is the pH of the resulting solution made by mixing 5 mL of 0.2178 M HCl and 15 mL of 0.1156 M NH₄OH? K_b = 1.8 x 10^-5 for NH₄OH.

Solution:

If an acid is mixed with a base, then the resulting product is a salt. If you mix a strong acid with a strong base, then the resulting product is a neutral salt. If you mix a weak acid with a weak base, then the resulting product is a neutral salt. However, if you mix a strong acid with a weak base, then the resulting product is an acidic salt. If you mix a weak acid with a strong base, then the resulting product is a basic salt. 

The pH of any neutral salt is usually equal to 7. The pH of an acidic salt is less than 7 and the pH of a basic salt is greater than 7. 

In the given problem, if HCl is mixed with NH₄OH, then the resulting product is
 

Since NH₄Cl is an acidic salt, then the pH is less than 7. From the given amount and concentration of the reactants, let's see if the resulting solution is acidic or basic as follows
 

 

 

Since the number of moles of NH₄OH is greater than with the number of moles of HCl, then the resulting mixture is basic. Well, let's see.

Hence, the resulting concentration of NH₄OH is
      

 

 

Since NH₄OH is a weak base, then its ionization is written as follows
 

The concentration of [OH^-] is 

Since the value of K_b is very small, then we can neglect x at the denominator as follows 

The pH of NH₄OH is

The resulting concentration of NH₄Cl which is their product is

The ionization of NH₄Cl is written as follows
 

 

 

Since [H⁺] is the product of the ionization process, then we need to convert K_b into K_a as follows

The concentration of [H⁺] in NH₄Cl is

Since the value of K_a is very small, then we can neglect x at the denominator as follows 

 

The pH of NH₄Cl is

Therefore, the pH of the overall resulting solution is

Solving for pH of a Salt Solution

Category: Chemical Engineering Math

"Published in Newark, California, USA"

What is the pH of 0.256 M NH₄Cl? K_b = 1.8 x 10^-5 for NH₄OH.

Solution:

The given salt, which is NH₄Cl is an acidic salt because it is a product of weak base and strong acid and HCl is a strong acid. In this case, the pH of an acidic salt is less than 7. The pH of a neutral salt is always equal to 7. Neutral salt is a product of strong acid and strong base or a product of weak acid and weak base. 

Let's consider the ionization of NH₄Cl as follows
 

 

 

Since [H⁺] is a product during the ionization process, then we have to get K_a from K_b as follows

The K_a of the ionization process of ammonium ion is 

If x is the concentration of [H⁺], then it follows that
 

If the concentration of NH₄Cl is 0.256 M, then it follows that
 

Hence, the concentration of [H⁺] during the ionization process is
 

 

Since the value of K_a is very small, then we can neglect x at the denominator as follows
 

 

 

 

 

Therefore, the pH of 0.256 M NH₄Cl is 
 

 

 

 

Solving for pH of a Buffer Solution, 3

Category: Chemical Engineering Math

"Published in Newark, California, USA"

What mass in grams of NaC₂H₃O₂ must be dissolved with 500 mL of 0.1 M acetic acid to make a buffer solution of pH = 5? K_a = 1.8 x 10^-5 for acetic acid.

Solution:

Buffer solution is a solution which is a mixture of weak acid and its salt or a mixture of weak base and its salt. The purpose of making a buffer solution is to prevent the rapid change of pH if acid or base is added to the solution. In this problem, the buffer solution is a mixture of acetic acid and sodium acetate. 

Let's consider the ionization of acetic acid as follows

The ionization constant for acetic acid is

If pH = 5 for the buffer solution, then the concentration of [H⁺] is

From the ionization of acetic acid, it follows that

Let's consider the ionization of sodium acetate as follows

If x is the concentration of sodium acetate, then it follows that

Again, let's consider the ionization constant of acetic acid as follows
 

 

If sodium acetate is added to acetic acid solution, then the concentration of acetate ion will increase as follows
 

 

Since 1 x 10^-5 is a very small number, then we can neglect it at the above equation as follows
 

 

 

Let's assume that the change of the volume of acetic acid is negligible if sodium acetate is added, then the volume of solution is also equal to 500 mL. 

Therefore, the weight of sodium acetate to be added with 500 mL of acetic acid is