122. Magesium fluoride, MgF 2 (aq) has a molar solubility of 2.7 10 3 mol/l. Use this iformatio to determie the K sp value for the solid. You eed to calculate the K sp for magesium fluoride, MgF 2. You kow the solubility of magesium fluoride is 2.7 10 3 mol/l. Write a balaced equatio for the solubility MgF 2 (s) Mg 2+ (aq) + 2F (aq) equilibrium. Write the expressio for the K sp. K sp = [Mg 2+ ][F ] 2 Determie the cocetratio of each io. 1 mol of MgF 2 (s) ioizes to give 1 mol of Mg 2+ ad 2 mol F io. Substitute the io cocetratios ito the K sp expressio ad solve. [Mg 2+ ] = [MgF 2 ] = 2.7 10 3 mol/l [F ] = 2[MgF 2 ] = 2 2.7 10 3 mol/l = 5.4 10 6 mol/l K sp = [Mg 2+ ][F ] 2 = (2.7 10 )(5.4 10 ) 2 = 7.9 10 8 The calculated value of K sp has the correct umber of sigificat digits that agrees with the give data. The small value is expected for a compoud of low solubility. 316 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
123. Silver sulfide, Ag 2 S(aq) has a K sp value that is equal to 5.6 10 49. What is the molar solubility of the solid? You eed to calculate the molar solubility of silver sulfide, Ag 2 S(s). You kow the K sp for silver sulfide is 5.6 10 49. Write the balaced equatio for the Write the solubility product equatio. Let x represet the molar solubility of Ag 2 S(s). Summarize the solubility of the Ag + (aq) ad S 2 (aq) ios at equilibrium i a ICE table. Ag 2 S(s) 2Ag + (aq) + S 2 (aq) K sp = [Ag + ] 2 [S 2 ] See the ICE table below. Ag 2 S(s) 2Ag + (aq) + S 2 (aq) [Ag + ] (mol/l) [S 2 ] (mol/l) I C x +2x +x E 2x x Write the expressio for the K sp ad solve for x. K sp = [Ag + ] 2 [S 2 ] 5.6 10 49 = (2x) 2 (x) = 4x 3 x = 5.2 10 17 mol/l The solubility is low as expected for a compoud with the give K sp. The aswer correctly shows 2 sigificat digits. Uit 4 Part B MHR 317
124. The K sp value for lead(ii) bromide, PbBr 2 (aq), is 6.6 10 6. What is the solubility of the solid? You eed to calculate the solubility of lead(ii) bromide(s) i g/l. You kow the K sp for lead(ii) bromide is 6.6 10 6. Write the balaced equatio for the Write the solubility product equatio. Let x represet the molar solubility of PbBr 2 (s). Summarize the solubility of the Pb 2+ (aq) ad Br (aq) ios at equilibrium i a ICE table. PbBr 2 (s) Pb 2+ (aq) + 2Br (aq) K sp = [Pb 2+ ][Br ] 2 See the ICE table below. PbBr 2 (s) Pb 2+ (aq) + 2Br (aq) [Pb 2+ ] (mol/l) [Br ] (mol/l) I C x +x +2x E x 2x Write the expressio for the K sp expressio ad solve for x. Determie the molar mass, M, of PbBr 2 ad express the solubility i g/l (m = M). K sp = [Pb 2+ ][Br ] 2 x 2 6 6.6 10 x 2 3 4x 2 x 1.18 10 mol/l 2 1.18 10 mol /L 367.00 g/ mol 4.3 g/l The solubility is low as expected for a compoud with the give K sp. The aswer correctly shows 2 sigificat digits. 318 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
125. Which solid ca have more mass ioize ito 1.00 L of solutio: silver chloride, AgCl(s), or copper(i) chloride, CuCl(s)? You eed to compare the solubility of silver chloride ad copper(i) chloride i g/l. From Appedix B, the K sp for silver chloride is 1.77 10 10 ad for copper(i) chloride is1.72 10 9. Write the balaced equatio for the solubility equilibrium of silver chloride. Write the solubility product equatio. Let x represet the molar solubility of AgCl(s). Write the expressio for the K sp. Let x represet the cocetratios of Ag + ad Cl. Solve for x. Determie the molar mass, M, of AgCl ad express the solubility i g/l (m = M). Repeat these steps for CuCl(s) ad determie which will have the greater mass dissolved i 1 L of solutio. AgCl(s) Ag + (aq) + Cl (aq) K sp = [Ag + ][Cl ] Ksp Ag Cl 10 1.77 10 ( x)( x) 2 x 10 1.77 10 x 1.33 10 mol/l 1.33 10 mol /L 143.32 g/ mol 3 1.91 10 g/l CuCl(s) Cu + (aq) + Cl (aq) Ksp Cu Cl 9 1.72 10 ( x)( x) x 4.15 10 mol/l 4.15 10 mol /L 99.00 g/ mol 3 4.11 10 g/l More CuCl(s) will dissolve per litre of solutio. The solubilities are expressed correctly to 3 sigificat digits ad the aswer is reasoable. Uit 4 Part B MHR 319
126. The solubility of ickel(ii) phosphate, Ni 3 (PO 4 ) 2 (aq), is 7.8 10 g/l. Determie the solubility-product costat for this solid. You eed to calculate the K sp for ickel(ii) phosphate, Ni 3 (PO 4 ) 2 (aq). You kow the solubility of Ni 3 (PO 4 ) 2 (aq) is 7.8 10 g/l. Use the molar mass, M, of Ni 3 (PO 4 ) 2 ad m the formula to express the M solubility i mol/l. Write a balaced chemical equatio for the Determie the cocetratio of each io. 1 mol of Ni 3 (PO 4 ) 2 ioizes to give 3 mol of Ni 2+ (aq) ios ad 2 mol PO 3 4 (aq) ios. m M 5 7.810 g /L 366.01g /mol 7 2.13 10 mol/l Ni 3 (PO 4 ) 2 (s) 3Ni 2+ (aq) + 2PO 4 3 (aq) [Ni 2+ ] = 3[Ni 3 (PO 4 ) 2 ] = 3 2.13 10 7 mol/l = 6.39 10 7 mol/l [PO 4 3 ] = 2[Ni 3 (PO 4 ) 2 ] = 2 2.13 10 7 mol/l = 4.26 10 7 mol/l Write the expressio for the K sp. K sp = [Ni 2+ ] 3 [PO 4 3 ] 2 Substitute the io cocetratios ito the K sp expressio ad solve. K sp = [Ni 2+ ] 3 [PO 4 3 ] 2 = (6.39 10 7 ) 3 (4.26 10 7 ) 2 = 4.74 10 32 The calculated value of K sp has the correct umber of sigificat digits that agrees with the give data. The small value for K sp is expected for a compoud of low solubility. 320 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
127. The solubility of strotium fluoride, SrF 2 (aq), is 12.2 mg/100 ml. What is K sp for this solid? You eed to calculate the K sp for strotium fluoride, SrF 2 (s). You kow the solubility of strotium fluoride, SrF 2 (aq), is 12.2 mg/100 ml. Use the molar mass, M, of SrF 2 ad the m formula to express the solubility i M mol/l. Write a balaced equatio for the Determie the cocetratio of each io. Oe mol of SrF 2 ioizes to give oe mol of Sr 2+ (aq) io ad two mol F (aq) ios. Write the expressio for the K sp. Substitute the io cocetratios ito the K sp expressio ad solve. 1 g 12.2 mg 1000 mg 1 L 100 ml 1000 ml = 0.122 g/l 0.122 g /L molar solubility = 125.62 g /mol = 9.71 10 4 mol/l SrF 2 (s) Sr 2+ (aq) + 2F (aq) [Sr 2+ ] = [SrF 2 ] = 9.71 10 4 mol/l [F ] = 2[SrF 2 ] = 2 2.71 10 4 mol/l = 1.94 10 3 mol/l K sp = [Sr 2+ ][F ] 2 = (9.71 10 4 )(1.94 10 3 ) 2 = 3.65 10 9 The calculated value of K sp has the correct umber of sigificat digits that agrees with the give data. The small value is expected for a compoud of low solubility. Uit 4 Part B MHR 321
128. Will a precipitate form if 1.00 ml of a 0.100 mol/l silver itrate solutio, AgNO 3 (aq), is added to 1.00 L of a 1.00 10 mol/l solutio of sodium chloride, NaCl(aq)? Show your calculatios. You eed to determie whether a precipitate will form. You kow that 1.00 ml of a 0.100 mol/l silver itrate solutio, AgNO 3 (aq), is added to 1.00 L of a 1.00 10 mol/l solutio of sodium chloride, NaCl(aq). The oly possible precipitate that ca form is silver chloride, AgCl(s). From Appedix B, you kow the K sp for silver chloride is 1.77 10 10. Write the balaced equatio for the Use the formula cv to determie the amout i moles,, of Ag + (aq) ad Cl (aq) i the origial solutio. AgCl(s) Ag + (aq) + Cl (aq) + Ag cv 0.100 mol/ L 0.001 L 4 1.00 10 mol Determie the total volume, V, after mixig the two solutios. Use the formula c to calculate the V molar cocetratios, c, of Ag + (aq) ad Cl (aq). Cl cv 1.00 10 mol/ L 1.00 L 1.00 10 mol V = 1.00 L + 1.00 ml = 1.00 L + 0.001 L = 1.001 L Ag V 1.001 L 4 1.00 10 mol 9.99 10 mol/l Cl V 1.001 L 5 1.00 10 mol 9.99 10 mol/l 322 MHR Chemistry 12 Solutios Maual 978-0-07-106042-4
Write the expressio for the K sp for AgCl. Q sp = [Ag + ][Cl ] Use the molar cocetratios calculated after mixig to determie a trial value Q sp. If Q sp > K sp, a precipitate of AgCl(s) forms. Q sp = [Ag + ][Cl ] = 9.99 10 9.99 10 = 9.98 10 10 K sp for AgCl is 1.77 10 10. Q sp > K sp A precipitate will form. The calculatios seem reasoable ad the correct umber of sigificat digits has bee used. Uit 4 Part B MHR 323