Question

What is the correct relationship between the $pHs$ of isomolar solutions of sodium oxide $(pH_{1})$, sodium sulphide $(pH_{2}),$ sodium selenide $(pH_{3})$ and sodium telluride $(pH_{4})$?

• A. $pH_{1} > pH_{2} = pH_{3} > pH_{4}$
• B. $pH_{1} < pH_{2} < pH_{3} < pH_{4}$
• C. $pH_{1} < pH_{2} < pH_{3} = pH_{4}$
• D. $pH_{1} > pH_{2} > pH_{3} > pH_{4}$

Answer 1

Answer: (D)

$pH_{1} > pH_{2} > pH_{3} > pH_{4}$

Answer 2

Order of acidic strength is

$H_{2}Te > H_{2}Se > H_{2}S > H_{2}O$

$Na_{2}O$is a salt of NaOH$+ H_{2}O$and $H_{2}O$is least acidic among given acids hence $pH$in this case will be $\max^{m}{}$.