(a) Explain briefly the observations in each of the following processes:
(i) when carbon (IV) oxide is bubbled through lime water, It turns milky but the milkiness disappears when the gas is bubbled for a long time.
(ii) A precipitate of calcium hydroxide is insoluble in excess sodium hydroxide solution whereas that of lead (ii) hydroxide is soluble
(b)(i) What is a primary standard solution?
(ii) Calculate the mass of sodium trioxocarbonate (V) required to prepare \(250\ \text{cm}^3\) of \(0.15\text{gmoldm}^3\) solution hydroxide is soluble. [Na = 23.0; O = 16.0; C = 12.0]
(c) Name one gas that can be collected by
(i) upward displacement of air
(ii) downward displacement of air
(a)(i) CO2 through lime water: at first, insoluble white calcium trioxocarbonate(IV) is formed, making the lime water milky:
\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]
On continued bubbling, excess CO2 converts the insoluble CaCO3 into soluble calcium hydrogentrioxocarbonate(IV), so the milkiness disappears (the solution becomes clear):
\[ \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \to \text{Ca(HCO}_3)_2 \]
(a)(ii) Calcium hydroxide is a base but not amphoteric, so it does not react with (dissolve in) excess NaOH and remains as a precipitate. Lead(II) hydroxide is amphoteric; it reacts with excess NaOH to form the soluble plumbite (tetrahydroxoplumbate(II)) ion, so it dissolves:
\[ \text{Pb(OH)}_2 + 2\text{OH}^- \to \text{Pb(OH)}_4^{2-} \]
(b)(i) Primary standard solution: a solution of a substance of known high purity, stable in air, of accurately known concentration prepared by weighing the solute directly (it does not absorb moisture or CO2 and has a high, well defined molar mass).
(b)(ii) Mass of Na2CO3 for 250 cm3 of 0.15 mol dm-3:
Molar mass Na2CO3 \(= (2\times23) + 12 + (3\times16) = 106\ \text{g mol}^{-1}\).
\[ \text{moles} = 0.15 \times \frac{250}{1000} = 0.0375\ \text{mol} \]
\[ \text{mass} = 0.0375 \times 106 = 3.98\ \text{g} \]
(c) Gas collection:
- (i) By upward displacement of air (denser than air): carbon(IV) oxide, CO2 (or chlorine, HCl).
- (ii) By downward displacement of air (less dense than air): ammonia, NH3 (or hydrogen).
(a)(i) CO2 through lime water: at first, insoluble white calcium trioxocarbonate(IV) is formed, making the lime water milky:
\[ \text{Ca(OH)}_2 + \text{CO}_2 \to \text{CaCO}_3\downarrow + \text{H}_2\text{O} \]
On continued bubbling, excess CO2 converts the insoluble CaCO3 into soluble calcium hydrogentrioxocarbonate(IV), so the milkiness disappears (the solution becomes clear):
\[ \text{CaCO}_3 + \text{CO}_2 + \text{H}_2\text{O} \to \text{Ca(HCO}_3)_2 \]
(a)(ii) Calcium hydroxide is a base but not amphoteric, so it does not react with (dissolve in) excess NaOH and remains as a precipitate. Lead(II) hydroxide is amphoteric; it reacts with excess NaOH to form the soluble plumbite (tetrahydroxoplumbate(II)) ion, so it dissolves:
\[ \text{Pb(OH)}_2 + 2\text{OH}^- \to \text{Pb(OH)}_4^{2-} \]
(b)(i) Primary standard solution: a solution of a substance of known high purity, stable in air, of accurately known concentration prepared by weighing the solute directly (it does not absorb moisture or CO2 and has a high, well defined molar mass).
(b)(ii) Mass of Na2CO3 for 250 cm3 of 0.15 mol dm-3:
Molar mass Na2CO3 \(= (2\times23) + 12 + (3\times16) = 106\ \text{g mol}^{-1}\).
\[ \text{moles} = 0.15 \times \frac{250}{1000} = 0.0375\ \text{mol} \]
\[ \text{mass} = 0.0375 \times 106 = 3.98\ \text{g} \]
(c) Gas collection:
- (i) By upward displacement of air (denser than air): carbon(IV) oxide, CO2 (or chlorine, HCl).
- (ii) By downward displacement of air (less dense than air): ammonia, NH3 (or hydrogen).
