Calculate the solubility of LaF3 in grams per liter in pure water. Calculate the

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(a) Lanthanum fluoride dissolves according to LaF3(s) ? M³?(aq) + 3 F?(aq) So the ionic molarities in a saturated solution satisfy the relation: Ksp = [La³?]·[F?]³ Let x be the molar solubility. When you dissolve x moles of LaF3 per L of pure water you get a just saturated solution in which the each molecule of LaF3 is dissolved into one lanthanum ion -ions and three fluoride ion. So the ionic molarities in that solutions are: [La³?] = x [F?] = 3·x Substitute equilibrium equation: Ksp = x·(3·x)³ = 27·x4 => x = ?( Ksp/27 ) = ?( 2.0×10?¹? / 27 ) = 9.28×10?6 mol/L Multiply by the molar mass of Lanthanum fluoride and you get the solubility: s = x · M = 9.28×10?6 mol/L · 195.9 g/mol = 1.82×10?³ g/L (b) The solution already contains fluoride ions at a concentration [F?]0 = 0.01mol/L. So the ionic molarities in a saturated solution from by adding x moles to the 0.051M KF solution are: [La³?] = x [F?] = [F?]0 + 3·x = 0.01+ 3·x Since x is expected to be small compared to 0.051 (due to common effect is should be smaller than x in part (a)), we can approximate [F?] ˜ [F?]0 = 0.01 Hence, Ksp = x·[F?]0³ => x = Ksp/[F?]0³ = 2.0×10?¹? / (0.01)³ = 20×10?¹5 mol/L => s = x · M = 20×10?¹5 mol/L · 195.9 g/mol = 39×10?¹³ g/L (c) Her the solution already contains lanthanum ions at a concentration [La³?]0 = 0.045mol/L. So the ionic molarities in a saturated solution from by adding x moles to the 0.051M LaCl3 solution are: [La³?] = [La³?]0 + x = 0.045+ x [F?] = 3·x Due to small magnitude of x we can approximate: [La³?] ˜ [La³?]0 = 0.045 Hence, Ksp = [La³?]0·(3·x)³ = 27·[La³?]0·x³ => x = ?( Ksp/(27·[La³?]0) ) = ?( 2.0×10?¹? / /(27 x 0.045) ) = 2.5425×10?7 mol/L => s = x · M = 2.5425×10?7 mol/L · 195.9 g/mol = 4.98375×10?6 g/L

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