A) \[CuS{{O}_{4}}<NaCN<NaCI={{H}_{2}}O\]
B) \[CuS{{O}_{4}}>NaCN={{H}_{2}}\text{O}>NaCl\]
C) \[CuS{{O}_{4}}={{H}_{2}}O>NaCN>NaCl\]
D) \[CuS{{O}_{4}}<NaCl={{H}_{2}}O<NaCN\]
Correct Answer: D
Solution :
\[\text{CuS}{{\text{O}}_{\text{4}}}\]is a salt of weak base \[[CO{{(OH)}_{2}}]\] and strong acid \[({{H}_{2}}S{{O}_{4}}),\] hence it will be acidic in nature. Therefore, pH of its solution must be less than 7. \[\text{NaCl}\] is a salt of strong base \[\text{(NaOH)}\] and strong acid \[\text{(HCl),}\] hence it will be neutral in nature. Therefore pH of its solution will be 7. Similarly, \[{{\text{H}}_{\text{2}}}\text{O}\]is neutral itself in nature, therefore its pH also will be 7. On the other hand, \[\text{NaCN}\] is a salt of strong base \[(\text{NaOH})\] and weak acid \[\text{(HCN),}\] so its solution must be basic in nature showing pH greater than 7. Hence, correct order of pH is as follows : \[CuS{{O}_{4}}<NaCl={{H}_{2}}O<NaCN.\]
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