#### Look at Figure 4.1 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 the same change? (d) How can a solution of lime water be prepared in the laboratory?

(a) Calcium hydroxide or Limewater $(Ca(OH)_2)$ taken in tube B turns milky when carbon dioxide passes through it as it forms Calcium carbonate $(CaCO_3)$

$Ca(OH)_2 (aq) + CO_2 (g) \rightarrow CaCO_3 (s) + H_2O (l)$

(b) Test Tube A

Sodium Carbonate reacts with ethanoic acid to form sodium ethanoate, carbon dioxide and water

$Na_2CO_3 + CH_3COOH \rightarrow CH_3COONa + CO_2 + H_2O$

Test Tube B

Calcium hydroxide turns milky white due to formation of Calcium carbonate (CaCO3) when Carbon dioxide is passed through it.

$Ca(OH)_2 (aq) + CO_2 (g) \rightarrow CaCO_3 (s) + H_2O (l)$

(c) No change will take place when we take ethanol instead of ethanoic acid.

$CH_3CH_2OH + Na_2CO_3 \rightarrow \text{ NO REACTION}$

(d) Take a small piece of Calcium oxide in a test tube. Add water to it, part of it gets dissolved but majority of it gets suspended. So pass the complete solution into a filter paper and collect it in a test tube and this solution is lime water.

$CaO + H_2O \rightarrow Ca(OH)_2$