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. $pH_1 > pH_2 \approx pH_3 > pH_4$
• 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

The Correct Option is 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 $Na _{2} O +2 H _{2} O \longrightarrow \underset{\text{strong base }}{2 NaOH} + \underset{\text{Water}}{H _{2} O}$ $Na _{2} S +2 H _{2} O \longrightarrow \underset{\text{strong base}}{2{ NaOH }}+ \underset{\text{weak acid }}{H _{2} S}$ $Na _{2} Se +2 H _{2} O \longrightarrow \underset{\text{strong base}}{2NaOH} + \underset{\text{weak acid}}{H _{2} Se}$ $Na _{2} Te +2 H _{2} O \longrightarrow \underset{\text{strong base}}{2 NaOH }+ \underset{\text{weak acid}}{H _{2} Te}$ Order of neutralisation of $NaOH$ $H _{2} Te > H _{2} Se > H _{2} S > H _{2} O$ Order of acidic strength $H _{2} Te > H _{2} Se > H _{2} S > H _{2} O$ 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 $)$.