A) \[N{{a}_{2}}S{{O}_{4}}\]
B) \[BiOCl\]
C) \[Pb(OH)Cl\]
D) \[N{{a}_{2}}HP{{O}_{4}}\]
Correct Answer: D
Solution :
\[N{{a}_{2}}HP{{O}_{4}}\,\to \,N{{a}_{2}}PO_{4}^{-}+{{H}^{+}}\] It can give \[{{H}^{+}}\] ion in solution.
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