
What is the colour of barium sulphate precipitate in the reaction of barium chloride and sodium sulphate?
Answer
405.3k+ views
Hint: Barium sulphate is an ionic compound consisting of barium and sulphate ions. Barium is an alkaline earth metal with a valency. Coloured salts are mainly formed by the transition metal ions that can show variable oxidation states and their electrons show d-d transition.
Complete answer:
The reaction between barium chloride and sodium sulphate is an example of a double displacement reaction where the anions and cations get interchanged giving rise to two new salts. The barium ion gets attracted to sulphate ions leaving behind chloride ions that form bonds with the free sodium ions. The reaction can be written as follows:
\[BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) \downarrow + NaCl(aq)\]
The sodium chloride formed in the product is highly soluble and remains dissolved in the solution while the barium sulphate precipitates out as a white solid.
The precipitation observed in the case of barium sulphate is a consequence of its low solubility. Both the barium as well as sulphate ions are large in size and therefore less soluble in water.
The precipitate of barium sulphate is white in colour as the salt does not release any free electrons that undergo transitions in the visible energy range.
Note:
Solubility is determined by both lattice energy as well as hydration enthalpy. Smaller ions tend to have hydration enthalpy that excess lattice energy making the ionic solids more soluble. The presence of both the large sized ions in barium sulphate leads to a higher lattice energy than the hydration enthalpy making the salt less soluble and easy to precipitate.
Complete answer:
The reaction between barium chloride and sodium sulphate is an example of a double displacement reaction where the anions and cations get interchanged giving rise to two new salts. The barium ion gets attracted to sulphate ions leaving behind chloride ions that form bonds with the free sodium ions. The reaction can be written as follows:
\[BaC{l_2}(aq) + N{a_2}S{O_4}(aq) \to BaS{O_4}(s) \downarrow + NaCl(aq)\]
The sodium chloride formed in the product is highly soluble and remains dissolved in the solution while the barium sulphate precipitates out as a white solid.
The precipitation observed in the case of barium sulphate is a consequence of its low solubility. Both the barium as well as sulphate ions are large in size and therefore less soluble in water.
The precipitate of barium sulphate is white in colour as the salt does not release any free electrons that undergo transitions in the visible energy range.
Note:
Solubility is determined by both lattice energy as well as hydration enthalpy. Smaller ions tend to have hydration enthalpy that excess lattice energy making the ionic solids more soluble. The presence of both the large sized ions in barium sulphate leads to a higher lattice energy than the hydration enthalpy making the salt less soluble and easy to precipitate.
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