A) \[Ca{{(OH)}_{2}}\]
B) \[N{{a}_{2}}C{{O}_{3}}\]
C) \[N{{a}_{2}}C{{O}_{3}}+Ca{{(OH)}_{2}}\]
D) None of these
Correct Answer: C
Solution :
\[Ca{{(OH)}_{2}}\] removes the permanent hardness due to \[M{{g}^{2+}}\] ion, but it produces \[C{{a}^{2+}}\] ions which are removed by \[N{{a}_{2}}C{{O}_{3}}\]. \[M{{g}^{2+}}+Ca{{(OH)}_{2}}\to Mg{{(OH)}_{2}}\downarrow +C{{a}^{2+}}\] \[C{{a}^{2+}}+N{{a}_{2}}C{{O}_{3}}\to CaC{{O}_{3}}\downarrow +2N{{a}^{+}}\] \[Ca{{(OH)}_{2}}\] or \[N{{a}_{2}}C{{O}_{3}}\] alone cannot remove the permanent hardness.
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