question_answer

What is the correct relationship between the pHs of isomolar solutions of sodium oxide \[(p{{H}_{1}}),\] sodium sulphide \[(p{{H}_{2}})\] sodium selenide \[(p{{H}_{3}})\] and sodium telluride \[(p{{H}_{4}})\]?    [AIPMT (S) 2005]

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)  \[p{{H}_{1}}>p{{H}_{2}}>p{{H}_{3}}>p{{H}_{4}}\]

Correct Answer: A

Solution :

The correct order of pH of isomolar solution in sodium oxide (pH1) sodium sulphide (pH2), sodium selenide (pH3) and sodium telluride (pH4) is pH1 > pH2 > pH3 > pH4 because in aqueous solution, they are hydrolysed as follows.

Order of acidic strength

Order of neutralisation of NaOH

Hence, their aqueous solutions have the following order of basic character due to neutralisation of NaOH with

H2O, H2S, H2Se and H2Te.

Na2O > Na2S > Na2Se > Na2Te

( pH of basic solution is higher than acidic or least basic solution)