Rahul added dilute hydrochloric acid to solid sodium hydrogen carbonate taken in a test tube. A gas was evolved which turned lime water milky. When the gas was passed through lime water in excess, the milkiness disappeared. Explain the chemical reactions involved by writing balanced chemical equations for both stages.

medium 3 marks Class 10 · Science · Acids, Bases and Salts

Answer

Stage 1:

NaHCO_{3} + HCl \to NaCl + H_{2} O + CO_{2} ↑

, and

Ca(OH)_{2} + CO_{2} \to CaCO_{3} ↓ + H_{2} O

. Stage 2:

CaCO_{3} + H_{2} O + CO_{2} \to Ca(HCO_{3})_{2}

Step-by-step solution

Step 1: Write the initial reaction between sodium hydrogen carbonate and dilute hydrochloric acid. $NaHCO_{3} (s) + HCl (a q) \to NaCl (a q) + H_{2} O (l) + CO_{2} (g)$ . The gas evolved is carbon dioxide ( $CO_{2}$ ).
Step 2: Explain why lime water turns milky. Carbon dioxide reacts with lime water (calcium hydroxide, $Ca(OH)_{2}$ ) to form an insoluble white precipitate of calcium carbonate ( $CaCO_{3}$ ). Equation: $Ca(OH)_{2} (a q) + CO_{2} (g) \to CaCO_{3} (s) + H_{2} O (l)$ .
Step 3: Explain why milkiness disappears in excess gas. When excess carbon dioxide is passed, insoluble calcium carbonate converts into soluble calcium hydrogen carbonate ( $Ca(HCO_{3})_{2}$ ). Equation: $CaCO_{3} (s) + H_{2} O (l) + CO_{2} (g) \to Ca(HCO_{3})_{2} (a q)$ .

Common mistake: Forgetting that excess carbon dioxide dissolves the white precipitate by forming soluble calcium hydrogen carbonate.

Concept tested: Reaction of carbonates with acids and lime water test