Look at the figure and answer the following questions. What change would you observe in the calcium hydroxide solution taken in tube B ?

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 .

Look at the figure and answer the following question. Write the reaction involved in test tubes A and B respectively.

Test tube A : Sodium carbonate reacts with ethanoic acid to give sodium ethanoate, carbon dioxide gas and water. Na 2 CO 3 + CH 3 COOH → CH 3 COONa + CO 2 + H 2 O Sodium Ethanoic Sodium Carbon carbonate acid ethanoate dioxide Test tube B : When carbon dioxide is passed through lime water, a milky-white precipitate of calcium carbonate is obtained. Ca OH 2 + CO 2 → CaCO 3 + H 2 O Calcium Carbon Calcium hydroxide dioxide carbonate ( lime water ) ( insoluble )

Look at the figure and answer the following questions. If ethanol is given instead of ethanoic acid, would you expect the same change?

If ethanol is taken instead of ethanoic acid, no change will occur because ethanol is a very weak acid and hence cannot decompose Na 2 CO 3 to give CO 2 gas. CH 3 CH 2 OH + Na 2 CO 3 → No reaction Ethanol

Look at the figure and answer the following questions. How can a solution of lime water be prepared in the laboratory?

When quick lime CaO is added to water in a test tube, some of it will dissolve to form calcium hydroxide while the majority of it remains suspended. When we filter the suspension, a clear solution is obtained called lime water. CaO S + H 2 O l → Suspension → Filter Quick lime Ca OH 2 aq Lime water