Question

What is the correct relationship between the pHs of isomolar solutions of sodium oxide $\left( pH _{1}\right)$, sodium sulphide $\left( pH _{2}\right)$, sodium selenide $\left( pH _{3}\right)$ and sodium telluride $\left( pH _{4}\right.$ )?

• A. $pH _{1}> pH _{2} \approx pH _{3}> pH _{4}$
• B. $pH _{1}< pH _{2}< pH _{3}< pH _{4}$
• C. $pH _{1}< pH _{2}< pH _{3} \approx pH _{4}$
• D. $pH _{1}> pH _{2}> pH _{3}> pH _{4}$

Answer 1

Answer: (D)

The correct order of $pH$ of isomolar solution in sodium oxide $\left( pH _{1}\right)$ sodium sulphide $\left( pH _{2}\right)$ sodium selenide $\left( pH _{3}\right)$ and sodium telluride $\left( pH _{4}\right)$ is $pH _{1}> pH _{2}> pH _{3}> pH _{4}$ because in aqueous solution, they are hydrolysed as follows.
$Na _{2} O +2 H _{2} O \longrightarrow \underset{\text { Base }}{2 NaOH }+ H _{2} O$
$Na _{2} S +2 H _{2} O \longrightarrow \underset{\text{Strong base }}{2 NaOH} +\underset{\text{Weak acid}}{H _{2} S}$
$Na _{2} Se +2 H _{2} O \longrightarrow \underset{\text{Strong base}}{2 NaOH} +\underset{\text{ Weak acid}}{H _{2} Se}$
$Na _{2} Te +2 H _{2} O \longrightarrow \underset{\text{Strong base}}{2 NaOH} +\underset{\text{ Weak acid}}{H _{2} Te}$
Order of acidic strength
$H _{2} Te > H _{2} Se > H _{2} S > H _{2} O$
Order of neutralisation of $NaOH$
$H _{2} Te > H _{2} Se > H _{2} S > H _{2} O$
Hence, their aqueous solutions have the following order of basic character due to neutralisation of $NaOH$ with $H _{2} O , H _{2} S . H _{2} Se$ and $H _{2} Te$
$Na _{2} O > Na _{2} S > Na _{2} Se > Na _{2} Te$
$(\because pH$ of basic solution is higher than acidic or least basic solution.$)$