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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
$pH_1 \lt pH_2 \lt pH_3 \lt pH_4$
B
$pH_1 \gt pH_2 \gt pH_3 \gt pH_4$
C
$pH_1 \lt pH_2 \lt pH_3 \approx pH_4$
D
$pH_1 \gt pH_2 \approx pH_3 \gt pH_4$
Detailed Solution
These are salts of the strong base NaOH with the very weak acids $H_2O$, $H_2S$, $H_2Se$ and $H_2Te$. In water the anions are hydrolysed and give $OH^-$ ions, making the solutions basic.
The acid strength of the hydrides increases down the group: $H_2O \lt H_2S \lt H_2Se \lt H_2Te$, because the H-E bond becomes longer and weaker.
The weaker the acid, the stronger its conjugate base, so the basic strength of the anions decreases: $O^{2-} \gt S^{2-} \gt Se^{2-} \gt Te^{2-}$.
$Na_2O$ reacts completely with water to give NaOH ($O^{2-} + H_2O \rightarrow 2OH^-$), so its solution is the most basic.
A more strongly basic solution has a higher pH.
So $pH_1 \gt pH_2 \gt pH_3 \gt pH_4$.
The acid strength of the hydrides increases down the group: $H_2O \lt H_2S \lt H_2Se \lt H_2Te$, because the H-E bond becomes longer and weaker.
The weaker the acid, the stronger its conjugate base, so the basic strength of the anions decreases: $O^{2-} \gt S^{2-} \gt Se^{2-} \gt Te^{2-}$.
$Na_2O$ reacts completely with water to give NaOH ($O^{2-} + H_2O \rightarrow 2OH^-$), so its solution is the most basic.
A more strongly basic solution has a higher pH.
So $pH_1 \gt pH_2 \gt pH_3 \gt pH_4$.