question_answer 87)

  When water is added to compound (A) of calcium, solution of compound (B) is formed. When \[C{{O}_{2}}\] is passed into the solution, it turns milky due to formation of compound (C). If excess of \[C{{O}_{2}}\] is passed into the solution, milkiness disappears due to formation of compound (D). Identify the compounds (A), (B), (C) and (D). Explain why milkiness disappears in the last step?

Answer:

  The compound (A) is quick lime, \[CaO\]. This combines with and forms calcium hydroxide, \[Ca{{(OH)}_{2}}\]. \[\underset{\underset{Calcium\,oxide}{\mathop{(A)}}\,}{\mathop{CaO}}\,+{{H}_{2}}O\to \underset{\begin{smallmatrix}  (B) \\  Calcium\,hydroxide \end{smallmatrix}}{\mathop{Ca{{(OH)}_{2}}}}\,\] When \[C{{O}_{2}}\] is passed through the solution having \[Ca{{(OH)}_{2}}\], the solution turns milky due to formation of calcium carbonate, \[CaC{{O}_{3}}\]. \[Ca{{(OH)}_{2}}+C{{O}_{2}}\to \underset{\begin{smallmatrix}  (C) \\  (Milkiness)  \\  Calcium\,carbonate \end{smallmatrix}}{\mathop{CaC{{O}_{3}}}}\,+{{H}_{2}}O\] When excess of \[C{{O}_{2}}\] is passed into milky solution, the milkiness disappears due to formation of calcium bicarbonate, \[Ca{{(HC{{O}_{3}})}_{2}}\] which is soluble. \[CaC{{O}_{3}}+{{H}_{2}}O+C{{O}_{2}}\to \underset{Calcium\,bicarbonte}{\mathop{Ca{{(HC{{O}_{3}})}_{2}}}}\,\]