Look at the figure and answer the following questions.

(a) What change would you observe in the calcium hydroxide solution taken in tube B ?
(b) Write the reaction involved in test tubes A and B respectively.
(c ) If ethanol is given instead of ethanoic acid, would you expect same change ?
(d) How can a solution of lime water be prepared in the laboratory ?

Explanation

(a) When `CO_(2)` is passed through calcium hydroxide solution (taken in test tube B), i.e., lime water, it turns milky due to the formation of insoluble calcium carbonate `(CaCO_(3))`.
(b) Test tube A
`underset("Sodium carbonate")(Na_(2)CO_(3))+underset("Ethanoic acid")(CH_(3)COOH)+underset("Carbon dioxide")(CO_(2))+H_(2)O`
(c) If ethanol taken instead of ethanoic acid, no change will because ethanol is a very weak acid and hence cannot decompose `Na_(2)CO_(3)` to give `CO_(2)` gas.
`underset("Ethanol")(CH_(3)CH_(2)OH)+N_(2)CO_(3)to"No reaction"`
(d) When quick lime (CaO) is added to water in a test tube. Some of it will dissolve to from calcium hydroxide (lime water) while majority of it remains suspended. Filter the solution. The clear solution thus obtained is called lime water.
`underset("Quick lime")(CaO(s))+H_(2)O(l)to"Suspension"overset("Filter")tounderset("Lime water")(Ca(OH)_(2)).(aq)`