A student measures the molar solubility of silver carbonate in a water solution to be 1.24 × 10⁻⁴ M. Based on her data, the solubility product constant for this compound is 7.63 × 10⁻¹² M³.
Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.
Silver carbonate dissociates as follows
Ag₂CO₃ (s) ⇄ 2Ag⁺ (aq) + CO₃⁻² (aq)
To find the solubility product constant use the expression
[tex]K_{sp} = [A^{+}]^{a} [B^{-}]^{b}[/tex]
where
[tex]K_{sp}[/tex] = Solubility product constant
A⁺ = cation in an aqueous solution
B⁻ = anion in an aqueous solution
a, b = concentration of a and b
Molar Solubility (S) = 1.24 × 10⁻⁴ M
Solubility, Ag⁺ = 2S
Solubility, CO₃⁻² = S
Put the value in above expression, we get
[tex]K_{sp} = [A^{+}]^{a} [B^{-}]^{b}[/tex]
= [Ag⁺]² [CO₃⁻²]¹
= [2S]² [S]¹
= 4S² × S
= 4S³
= 4 × ( 1.24 × 10⁻⁴ M)³
= 7.63 × 10⁻¹² M³
Thus from the above conclusion we can say that A student measures the molar solubility of silver carbonate in a water solution to be 1.24 × 10⁻⁴ M. Based on her data, the solubility product constant for this compound is 7.63 × 10⁻¹² M³.
Learn more about the Solubility Product Constant here: https://brainly.com/question/1419865
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Question: A student measures the molar solubility of silver carbonate in a water solution to be 1.24 × 10⁻⁴ M. Based on her data, the solubility product constant for this compound is ..........
