What is the correct relationship between the pH_s of isomolar solutions of sodium oxide (pH₁),sodium sulphide (pH₂),sodium selenide (pH₃) and sodium telluride (pH₄)?

Question

Find the solubility product of \( \mathrm{Ba(OH)_2} \)?
Where \( S = 1.73 \times 10^{-14} \).

Answer 1

To determine the correct relationship between the pH values of isomolar solutions of sodium oxide (pH₁), sodium sulphide (pH₂), sodium selenide (pH₃), and sodium telluride (pH₄), we must consider the nature of the reactions of these compounds in water.

These compounds are oxides and salts of strong bases and weak acids. When dissolved, they will produce basic solutions, contributing hydroxide ions (OH^-) to the solution. The degree of basicity, which affects the pH, depends on the strength of the accompanying weak acids.

Sodium oxide (Na₂O): Dissolves in water to form sodium hydroxide (NaOH), a strong base. The reaction is:
Na_2O + H_2O \rightarrow 2 NaOH. This solution will have the highest pH, as NaOH dissociates completely to give a highly alkaline solution.
Sodium sulphide (Na₂S): Dissolves to produce the weak acid anion H₂S, but NaOH is stronger here. This will result in a relatively high pH, though lower than Na₂O.
Sodium selenide (Na₂Se): Similar to Na₂S, but H₂Se is a stronger acid than H₂S, resulting in a lower pH compared to Na₂S.
Sodium telluride (Na₂Te): Analogous to the above compounds, H₂Te is an even stronger acid than H₂Se, thus giving the lowest pH among these compounds.

Based on the above explanations, the trend in basicity (and thus pH) of these solutions is:

pH₁ > pH₂ > pH₃ > pH₄.

Therefore, the correct relationship is that pH₁ is greater than pH₂, which is greater than pH₃, and finally greater than pH₄. This sequence corresponds to the strength of acid displacement in the hydracids of chalcogens (oxygen family) where, as we move from sulphur to tellurium, the acidic strength increases, thus reducing the basicity of solutions formed by the respective sodium compounds.

The correct option is: pH₁ > pH₂ > pH₃ > pH₄.

Answer 2

To determine the correct relationship between the pH values of isomolar solutions of sodium oxide (pH₁), sodium sulphide (pH₂), sodium selenide (pH₃), and sodium telluride (pH₄), we must consider the nature of the reactions of these compounds in water.

These compounds are oxides and salts of strong bases and weak acids. When dissolved, they will produce basic solutions, contributing hydroxide ions (OH^-) to the solution. The degree of basicity, which affects the pH, depends on the strength of the accompanying weak acids.

Sodium oxide (Na₂O): Dissolves in water to form sodium hydroxide (NaOH), a strong base. The reaction is:
Na_2O + H_2O \rightarrow 2 NaOH. This solution will have the highest pH, as NaOH dissociates completely to give a highly alkaline solution.
Sodium sulphide (Na₂S): Dissolves to produce the weak acid anion H₂S, but NaOH is stronger here. This will result in a relatively high pH, though lower than Na₂O.
Sodium selenide (Na₂Se): Similar to Na₂S, but H₂Se is a stronger acid than H₂S, resulting in a lower pH compared to Na₂S.
Sodium telluride (Na₂Te): Analogous to the above compounds, H₂Te is an even stronger acid than H₂Se, thus giving the lowest pH among these compounds.

Based on the above explanations, the trend in basicity (and thus pH) of these solutions is:

pH₁ > pH₂ > pH₃ > pH₄.

Therefore, the correct relationship is that pH₁ is greater than pH₂, which is greater than pH₃, and finally greater than pH₄. This sequence corresponds to the strength of acid displacement in the hydracids of chalcogens (oxygen family) where, as we move from sulphur to tellurium, the acidic strength increases, thus reducing the basicity of solutions formed by the respective sodium compounds.

The correct option is: pH₁ > pH₂ > pH₃ > pH₄.