Calculate the molar solubility of silver thiocyanate, AgSCN, in pure water and i

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Solution: As we know that Silver thiocyanate dissolves according to: AgSCN(s) ? Ag?(aq) + CN?(aq) we can write in equilibrium equation: Ksp = [Ag?]·[CN?] Hence the Solubility product constant for AgSCN is 1: So =>Ksp = 1.0×10?¹² Now: 1 Solubility in pure water. Let we assume that x be the molar solubility. Because each salt molecule dissolves to: one silver ion and on thiocyanate ion, So we conclude that adding x moles of AgCN per liter yields a saturated solution with ionic molarities: => [Ag?] = [CN?] = x Hence; Ksp = x·x = x² =>x = vKsp = v1.0×10?¹² = 1.0×10?6 (mol/L) Noq we have: 2 Solubility in 0.01M NaSCN Silver ion concentration is as in 1st part so: [Ag?] = x Since solution contains: 0.01M thiocyanate ions from dissolved NaSCN: => [CN?] = 0.01 + x Because x is expected to be small compared to 0.01 we may approximate: [CN?] ˜ 0.01 =>Ksp = x·[CN?] =>x = Ksp / [CN?] = 1.0×10?¹²/0.01 = 1.0×10?¹° mol/L Ans Please rate

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