Department of Chemistry University of Texas at Austin Name:____________________ Solubility Product Constant (Ksp) – Supplemental Worksheet KEY 1. What is the solubility, in moles/liter, of AgI if the Ksp = 8.5•10-‐12? AgI (s) à Ag+ (aq) + I– (aq) Ksp = [Ag+]1[ I–]1 We know our Ksp = 8.5•10-‐12 and we can insert this into our Ksp equation 8.5•10–12 = [Ag+]1[ I–]1 Since we have a 1:1 ion-‐ion dissociation of AgI, we can set x to the following then plug into our Ksp equation: x = [Ag+] = [ I–] 8.5•10–12 = x . x 8.5•10–12 = x2 2.93•10–6 = x x = 2.93•10–6 mol L-‐1 = [Ag+] = [ I–] SAgI = 2.93•10–6 mol L-‐1 2. If the solubility of Li2CO3 is 1.32 g/100 mL, what is its Ksp at room temperature? Li2CO3 (s) à 2Li+ (aq) + CO32– Ksp = [Li+]2[CO32–]1 2Li+ = 2x ; CO32– = x Ksp = (2x)2(x) = 4x3 Next, we need to determine the solubility in terms of mol/L. Since we have been given the solubility in terms of g/mL, we need to convert this. SLi2CO3 = 1.32 g Li2CO3 1 mol Li2CO3 1000 mL 100 mL 73.891g Li2CO3 1 L = 0.1786 mol L-‐1 Li2CO3 Revised CR 12/10/13 © LaBrake & Vanden Bout 2013 Department of Chemistry University of Texas at Austin Name:____________________ Since we have converted the solubility into mol/L we can insert this into our Ksp. The solubility is for one mole of Li2CO3 dissolving. For every one mole of the substance we observe one mole of CO32-‐ which we earlier called “x”. So the solubility of Li2CO3 is equal to x. Ksp = 4(0.1786 mol L-‐1)3 = 2.3 x 10–2 3. What is the solubility, in moles/liter, of HgBr2 if the Ksp = 6.2•10-‐9? HgBr2 (s) à Hg2+ (aq) + 2Br – (aq) Ksp= [Hg2+]1[Br –]1 2Br – = 2x ; Hg2+ = x Ksp= (2x)2(x) = 4x3 6.2•10–9 = 4x3 In order to find the solubility at this step, we must solve for x! 6.2•10–9 = 4x3 6.2•10–9 ÷ 4 = x3 3 6.2 • 10 −9 = x 4 x = 0.00157 = 1.157•10–3 mol L-‐1 € SHgBr2 = 1.157•10–3 mol L-‐1 4. If Cu3(AsO4)2 has a Ksp = 8.0•10-‐36, then what is the concentration of [Cu2+] in a saturated solution? Cu3(AsO4)2 (s) à 3Cu2+ (aq) + 2AsO43– Ksp = [Cu2+]3[AsO43–]2 3Cu2+ = 3x ; 2AsO43– = 2x Ksp = (3x)3(2x)2 = (27x3)(4x2) = 108x5 8.0•10–36 = 108x5 8.0•10–36 ÷ 108 = x5 Revised CR 12/10/13 © LaBrake & Vanden Bout 2013 Department of Chemistry University of Texas at Austin Name:____________________ 5 8.0 • 10 −36 = x 108 x = 3.75•10–8 [Cu2+€ ] = 3x = 3(3.75•10–8) = 1.13•10–7 mol L-‐1 5. How many moles of AgCl will dissolve in 250 mL of water if the Ksp = 1.7•10-‐10? First, we must find the solubility of AgCl. We can do so by the following: AgCl (s) à Ag+ (aq) + Cl– (aq) Ksp = [Ag+]1[Cl–]1 Since we have a 1:1 ion-‐ion dissociation of AgI, we can set x to the following then plug into our Ksp equation: x = [Ag+] = [Cl–] 1.7•10–10 = x . x 1.7•10–10 = x2 1.30•10–5 = x x = 1.30•10– 5= [Ag+] = [Cl–] SAgCl = 1.30•10– 5mol L–1 Second, we must find the amount of moles of AgCL in 0.250 L of water: nAgCl = SAgCl . VH2O nAgCl = (1.30•10– 5mol L–1)(0.250 L) = 3.25•10–6 moles of AgCl Revised CR 12/10/13 © LaBrake & Vanden Bout 2013

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