Question

What is the correct relationship between the pHs of isomolar solutions of sodium oxide $\left( {{\text{p}}{{\text{H}}_{\text{1}}}} \right)$, sodium sulphide $\left( {{\text{p}}{{\text{H}}_{\text{2}}}} \right)$, sodium selenide $\left( {{\text{p}}{{\text{H}}_{\text{3}}}} \right)$ and sodium telluride $\left( {{\text{p}}{{\text{H}}_{\text{4}}}} \right)$?
(A) ${\text{p}}{{\text{H}}_{\text{1}}} > {\text{p}}{{\text{H}}_{\text{2}}} = {\text{p}}{{\text{H}}_{\text{3}}} > {\text{p}}{{\text{H}}_{\text{4}}}$
(B)${\text{p}}{{\text{H}}_{\text{1}}} < {\text{p}}{{\text{H}}_{\text{2}}} < {\text{p}}{{\text{H}}_{\text{3}}} < {\text{p}}{{\text{H}}_{\text{4}}}$
(C) ${\text{p}}{{\text{H}}_{\text{1}}} < {\text{p}}{{\text{H}}_{\text{2}}} < {\text{p}}{{\text{H}}_{\text{3}}} = {\text{p}}{{\text{H}}_{\text{4}}}$
(D) ${\text{p}}{{\text{H}}_{\text{1}}} > {\text{p}}{{\text{H}}_{\text{2}}} > {\text{p}}{{\text{H}}_{\text{3}}} > {\text{p}}{{\text{H}}_{\text{4}}}$

Answer 1

We are given isomolar solutions of sodium salts of group 16 elements. We can determine the pH of isomolar solutions with respect to the basic character of the given sodium salts of group 16 elements. These salts undergo hydrolysis and form hydrides. The acidic character of hydrides can account for the basic character of the salts formed by neutralisation.

Complete step-by-step answer: We are given isomolar solutions of sodium salts of group 16 elements. The sodium salts are sodium oxide $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}} \right)$, sodium sulphide $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}} \right)$, sodium selenide $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{Se}}} \right)$ and sodium telluride $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{Te}}} \right)$.
Isomolar solutions meant that sodium oxide $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{O}}} \right)$, sodium sulphide $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}} \right)$, sodium selenide $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{Se}}} \right)$ and sodium telluride $\left( {{\text{N}}{{\text{a}}_{\text{2}}}{\text{Te}}} \right)$ equal molar concentration.
These salts undergo hydrolysis on reaction with water and produce acid and base. The reactions are as follows:
${\text{N}}{{\text{a}}_{\text{2}}}{\text{O}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2NaOH}} + {{\text{H}}_{\text{2}}}{\text{O}}$
${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2NaOH}} + {{\text{H}}_{\text{2}}}{\text{S}}$
${\text{N}}{{\text{a}}_{\text{2}}}{\text{Se}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2NaOH}} + {{\text{H}}_{\text{2}}}{\text{Se}}$
${\text{N}}{{\text{a}}_{\text{2}}}{\text{Te}} + {\text{2}}{{\text{H}}_{\text{2}}}{\text{O}} \to {\text{2NaOH}} + {{\text{H}}_{\text{2}}}{\text{Te}}$
In all the reactions, sodium hydroxide which is a strong base is formed. The hydrides of group 16 are elements formed.The acidic strength of the hydrides increases down the group. This is because down the group atomic size increases. As the atomic size increases the bond between metal and hydrogen lengthens and becomes weak. Thus, the proton can be easily donated and the acidic strength decreases.
Thus, the order of acidic strength of hydrides is as follows:
${{\text{H}}_{\text{2}}}{\text{O}} < {{\text{H}}_{\text{2}}}{\text{S}} < {{\text{H}}_{\text{2}}}{\text{Se}} < {{\text{H}}_{\text{2}}}{\text{Te}}$
Higher the acidic strength of the hydride, higher is the neutralization to form the salt. Thus, as the acidic strength of hydrides increases down the group, the basic strength of the salts decreases down the group. Thus, the order of basic strength of the salts is as follows:
${\text{N}}{{\text{a}}_{\text{2}}}{\text{O}} > {\text{N}}{{\text{a}}_{\text{2}}}{\text{S}} > {\text{N}}{{\text{a}}_{\text{2}}}{\text{Se}} > {\text{N}}{{\text{a}}_{\text{2}}}{\text{Te}}$
The basicity of the salt is directly proportional to the pH of the salt solution. Thus, as the basicity increases, the pH also increases. Thus, the order of pHs of isomolar solutions is as follows:
${\text{p}}{{\text{H}}_{\text{1}}} > {\text{p}}{{\text{H}}_{\text{2}}} > {\text{p}}{{\text{H}}_{\text{3}}} > {\text{p}}{{\text{H}}_{\text{4}}}$

Thus, the correct option is (D) ${\text{p}}{{\text{H}}_{\text{1}}} > {\text{p}}{{\text{H}}_{\text{2}}} > {\text{p}}{{\text{H}}_{\text{3}}} > {\text{p}}{{\text{H}}_{\text{4}}}$.

Note: If the pH of the solution is less than 7 then the solution is acidic in nature. If the pH of the solution is equal to 7 then the solution is neutral in nature. If the pH of the solution is more than 7 then the solution is basic in nature. Here, the pH is 11 thus, we can say that the solution is basic or alkaline in nature.