ABSE_pH of Salt Solutions Lesson

ABSE_pH of Salt Solutions

image of the Salt FlatsWe have classified chemicals as acids, bases, or salts. You probably have never thought of the pH of a salt solution. The reason is that most of them are neutral. However, not all salt solutions are neutral. All salt solutions contain a cation and anion. Sometimes one or both of these ions can actually make the pH of an aqueous solution a value other than neutral.  

Salts that Produce Neutral Solutions

To determine the pH range of a salt solution, we need to look at the cation and anion separately.

Cation

If you have a salt that contains a cation that is a metal of small charge, that cation does not affect the loss of protons by water molecules. These cations therefore do not affect the pH of a salt solution. A metal cation of small charge includes group IA and group IIA, except Ba2+.

For example:

Consider a salt that contains Na+. Na+ is a metal cation from group IA and does not change the pH of the solution.

Anion

If you have a salt that contains an anion that is the conjugate base of a strong acid, that anion does not have an affinity to react to form a molecule. These anions therefore do not affect the pH of a salt solution.

For example:

Consider a salt that contains Cl-. Cl- is the conjugate base of HCl, a strong acid. The Cl- ion does not change the pH of the solution.

Therefore, salts that contain the cations from group IA and group IIA (except Ba2+) and anions that are conjugate bases of a strong acids, they have a neutral pH.

For example, these following salts have a pH of 7:

NaCl, LiBr, KNO3, Na2SO4, KClO4

Salts that Produce Acidic or Basic Solutions

When some salts dissolve, the ions in solution do not remain in ion form. These ions can react with water. This can result in the formation of either H3O+ or OH-. This process is called hydrolysis. Ions that undergo hydrolysis produce products that can affect the pH of the solution.

So, which species of salt solutions undergo hydrolysis?

  • Cations that are conjugate acids of weak molecular bases are weak acids and will undergo hydrolysis.

Ex: The ammonium ion, NH4 +, is the conjugate acid of NH3, a weak base. It will undergo hydrolysis:    

NH4 + +   H2 ⇌ NH3   +   H3O+

  • Anions that are conjugate bases of weak acids are weak bases and will undergo hydrolysis.

Ex: The acetate ion, C2H3O2 -, is the conjugate base of HC2H3O2, a weak acid. It will undergo hydrolysis:    

C2H3O2 - +   H2O  ⇌ HC2H3O2   +   OH-

If a salt solution undergoes hydrolysis and forms H3O+ , the salt solution will be acidic.  

If a salt solution undergoes hydrolysis and forms OH-, the salt solution will be basic.

Determine if the following salt solution will be acidic, basic, or neutral. Write the reaction that justifies your answer. Ammonium chloride Start by identifying the ions. NH₄⁺ and Cl- We will look at the cation first. If the cation is from Group IA or IIA (except Ba²⁺) it will not affect the pH. If it is the conjugate acid of a molecular base it is a weak acid and will affect the pH. NH₄⁺ is not from group IA or IIA. It is the conjugate acid of NH₃, a molecular base. So, NH₄⁺ will affect the pH. Now, we will look at the anion. Write the conjugate acid of the anion. If the conjugate acid is strong, the anion will not affect the pH. If the conjugate acid is weak, the anion will affect the pH. The conjugate acid of Cl- is HCI. HCI is a strong acid, so the Cl- will not affect the pH of the salt solution. Now we can then write the hydrolysis equation for the ion determined to change the pH of the salt solution. NH₄⁺ + H₂O ↔️NH₃ + H₃O⁺ Since one of the products is H₃O⁺, this salt is acidic.

Here is another example, with fewer explanations given.

Determine if the following salt solution will be acidic, basic, or neutral. Write the reaction that justifies your answer. Sodium hypochlorite lons: Na⁺ and OCI- Cation: Na⁺ is from group IA, and does not change the pH. Anion: OCI- is the conjugate of HOCI. HOCI is a weak acid, and OCI- will affect the pH. Hydrolysis: OCI + H2O ↔️HOCI + OH- Since one of the products is OH, this salt is basic.

Note that you do not have to show all these steps when answering these type questions. This just shows you the thought process you must go through to find the answer.

In many salt solutions, both the cation and the anion can undergo hydrolysis and affect the pH of the solution. In these cases, you must compare the ka and kb to determine if the solution will be acidic, basic, or neutral. We must determine which reaction occurs to the greater extent. If the ka is larger than the kb, then the ka reaction favors the products more than does the kb reaction. The resulting solution will therefore contain more H3O+ and the pH will be less than 7 (acidic).   The reverse is true if ka is smaller than the kb.

  • If ka = kb, the salt solution will be neutral.
  • If ka > kb, the salt solution will be acidic
  • If kb > ka, the salt solution will be basic.

Will an ammonium fluoride solution be acidic, basic, or neutral? Start by writing the hydrolysis reaction for each ion. NH₄⁺+ H₂O ↔️NH₃ + H₃O⁺ F⁻+ H₂O ↔️HF + OH- We need to compare the ka of NH₄⁺ and kb of F⁻. Look these up on the ka chart in the back of your book. NH₄⁺+ H₂O ↔️NH₃ + H₃O⁺ Ka = 5.6 x 10⁻¹⁰ Often only the ka values are given. Therefore, you will need to use the ka of the conjugate acid and calculate kb. Ka of HF = 6.8 x 10⁻⁴ Ka X Kb = Kw 6.8 x 10⁻⁴ x kb = 1.0 x 10⁻¹⁴ F⁻+ H₂O↔️ HF + OH- kb= 1.5 x 10⁻¹¹ Ka > Kb, therefore, the salt solution is acidic.

Calculating the pH of Salt Solutions

We can determine if salt solutions are acidic, basic, or neutral by inspection, but in order to determine the exact pH we must know the exact concentration of H+.

Example Calculate the pH of a 0.10 M solution of sodium hypochlorite. To determine the pH, write the hydrolysis equation for the ion that will affect the pH. Write the kb expression (calculating kь from ka if necessary). k = [HOCI][OH]/[OCI] x²= 3.3 x 10⁻⁷ 3.3 x 10⁻⁷ (0.10 -x) The denominator can be assumed to be 0.10. x²/(0.10)-x) = 3.3 x 10⁻⁷ x = 1.82 x 10⁻⁴ = [OH-] pOH = -log[OH-]eq pOH = -log (1.82 x 10⁻⁴) pOH = 3.7 pH + pOH = 14 pH = 10.3

Remember to work on the module practice problems as you complete each section of content.  

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