Question

What is the correct relationship between the pH of isomolar solutions of sodium oxide $(p{{H}_{1}})$, sodium sulphide $(p{{H}_{2}})$, sodium selenide $(p{{H}_{3}})$, and sodium telluride $(p{{H}_{4}})$.
(A)- $p{{H}_{1}}>p{{H}_{2}}>p{{H}_{3}}>p{{H}_{4}}$
(B)- $p{{H}_{1}}>p{{H}_{2}}=p{{H}_{3}}>p{{H}_{4}}$
(C)- $p{{H}_{1}}$$<$$p{{H}_{2}}$$<$$p{{H}_{3}}$$<$$p{{H}_{4}}$
(D)- $p{{H}_{1}}$$<$$p{{H}_{2}}$$<$$p{{H}_{3}}$$=$$p{{H}_{4}}$

Answer 1

Hint: The pH of the isomolar solution, can be determined with respect to the basic character of the sodium salts of the Group 16 elements. These salts undergo hydrolysis, forming hydrides, whose acidic character can account for the basic character of the salt, formed through neutralisation.

Complete step by step solution:
It is given that the isomolar solutions are sodium salt of Group 16 elements, that is $N{{a}_{2}}O,\,N{{a}_{2}}S,\,N{{a}_{2}}Se,\,N{{a}_{2}}Te$ are present in equal molar concentration. The salts on hydrolysis with water, it produces an acid and base, as follows:
$N{{a}_{2}}O+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}O$
$N{{a}_{2}}S+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}S$
$N{{a}_{2}}Se+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}Se$
$N{{a}_{2}}Te+2{{H}_{2}}O\to 2NaOH+{{H}_{2}}Te$
So, the base formed, that is, sodium hydroxide in each reaction is a strong base. Then, the acidic strength of the hydrides of Group 16 elements is in the order ${{H}_{2}}O<{{H}_{2}}S<{{H}_{2}}Se<{{H}_{2}}Te$ .
That is, down the group, the acidic strength of the hydrides decreases, because with the increase in the atomic size, the bond between M-H lengthens. So, the proton can be easily donated as the M-H bond weakens down the group.
Now, in the reverse reaction of neutralisation, from the acidic strength of the hydrides, we can now obtain the order of ease of neutralisation of the sodium hydroxide with the hydrides. As more the acidity of the hydride, more will be the neutralisation to form the salt. Then, the order will be:
${{H}_{2}}O<{{H}_{2}}S<{{H}_{2}}Se<{{H}_{2}}Te$
So, as the acidic character of the hydride increases down the group, the basic character of the salt decreases down the group. Therefore, the order of basic character of the salts will be as follows:
$N{{a}_{2}}O>N{{a}_{2}}S>N{{a}_{2}}Se>N{{a}_{2}}Te$
Then, the basicity of the salt is directly proportional to the pH of the salt solution. As more the basic character, more will be the pH value.

Therefore, the order of pH of the given aqueous solution will be option (A)- $p{{H}_{1}}>p{{H}_{2}}>p{{H}_{3}}>p{{H}_{4}}$.

Note: When the value of pH is more than 7, then the solution is basic. Higher the value of pH, more basic is the solution.
Also, the hydrolysis and the neutralisation reaction are reversed.