{"id":14769,"date":"2022-11-20T10:00:07","date_gmt":"2022-11-20T04:30:07","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=14769"},"modified":"2022-11-21T10:18:19","modified_gmt":"2022-11-21T04:48:19","slug":"heat-of-precipitation","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/heat-of-precipitation\/","title":{"rendered":"What is heat of precipitation?"},"content":{"rendered":"
<\/p>\n
Heat of precipitation:<\/strong><\/p>\n Some examples of precipitation reactions are:<\/p>\n People also ask<\/strong><\/p>\n Aim:<\/strong> To determine the heat of precipitation of silver chloride. Results:<\/strong> Interpreting data: Discussion:<\/strong><\/p>\n Conclusion:<\/strong> 1.\u00a0<\/strong>50 cm3<\/sup> of 0.5 mol dm-3<\/sup> silver nitrate solution at 29.5\u00b0C is added to 50 cm3<\/sup> of 0.5 mol dm-3<\/sup> potassium chloride solution which is at a temperature of 28.5\u00b0C. The mixture is stirred and the highest temperature reached is 32.0\u00b0C.\u00a0Calculate the heat of precipitation for silver chloride. 2.<\/strong> The thermochemical equation for the precipitation of lead(II) sulphate is given below. 3.\u00a0<\/strong>When 50 cm3<\/sup> of 2.0 mol dm-3<\/sup> lead(II) nitrate solution is added to 50 cm3<\/sup> of 2.0 mol dm-3<\/sup>\u00a0potassium sulphate solution, there is an increase of 10\u00b0C in the temperature. (b)\u00a0<\/strong>The concentration of both reactants are halved. (c)\u00a0<\/strong>The concentration of potassium sulphate solution is halved. When 50 cm3<\/sup> of 2.0 mol dm-3<\/sup> lead(II) nitrate solution and 50 cm3<\/sup> of 1.0 mol dm-3<\/sup> potassium sulphate solution are used, (d)\u00a0<\/strong>50 cm3<\/sup> of 2.0 mol dm-3<\/sup> of potassium sulphate solution is replaced by 50 cm3<\/sup> of 2.0 mol dm-3<\/sup> sodium sulphate solution. What is heat of precipitation? Heat of precipitation: When two aqueous solutions are added together and a precipitate is formed, this reaction is called a precipitation reaction or double decomposition. This reaction is used to prepare insoluble salts. The heat given out in a precipitation reaction is called the heat of precipitation. The heat […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[84],"tags":[5831,5819,5826,5823,5821,5822,5820,5824,5825,5829,5828,5827,5813,5818,5830,5817],"yoast_head":"\n\n
\n
\nBaCl2<\/sub>(aq) + Na2<\/sub>SO4<\/sub>(aq) \u2192 BaSO4<\/sub>(s) + 2NaCl(aq)
\nBa2+<\/sup>(aq) + SO4<\/sub>2-<\/sup>(aq) \u2192 BaSO4<\/sub>(s)<\/li>\n
\nAgNO3<\/sub>(aq) + HCl(aq) \u2192 AgCl(s) + HNO3<\/sub>(aq)
\nAg+<\/sup>(aq) + Cl–<\/sup>(aq) \u2192\u00a0AgCl(s)<\/li>\n
\nPb(NO3<\/sub>)2<\/sub>(aq) + 2KI(aq) \u2192 PbI2\u00a0<\/sub>+ 2KNO3<\/sub>(aq)
\nPb2+<\/sup>(aq) + 2I–<\/sup>(aq) \u2192\u00a0PbI2<\/sub>(s)<\/li>\n<\/ul>\n\n
\n<\/a><\/li>\nHeat of precipitation of silver chloride experiment<\/strong><\/h2>\n
\nMaterials:<\/strong> 0.5 mol dm-3<\/sup> silver nitrate solution, 0.5 mol dm-3<\/sup> sodium chloride solution.
\nApparatus:<\/strong> Thermometer, plastic cups with covers, 50 cm3<\/sup> measuring cylinders.
\nCaution:
\n<\/strong>Add the two solutions together as quickly as possible.
\nStir the mixture throughout the activity.
\nSafety measures:<\/strong>
\nHandle the chemicals with caution. Wear eye protection.
\nSilver nitrate solution may cause dark stains on your skin and clothing.
\nProcedure:
\n
\n<\/strong><\/p>\n\n
\n<\/p>\n
\n
\n<\/strong><\/p>\n\n
\nAgNO3<\/sub>(aq) + NaCl(aq) \u2192\u00a0AgCl(s) + NaNO3<\/sub>(aq)<\/li>\n
\nAg+<\/sup>(aq) + Cl–<\/sup>(aq) \u2192\u00a0AgCl(s)<\/li>\n
\nAgNO3<\/sub>(aq) + NaCl(aq) \u2192\u00a0AgCl(s) + NaNO3<\/sub>(aq) \u00a0 \u00a0\u0394H = -0.1 y kJ mol-1<\/sup>
\nor
\nAg+<\/sup>(aq) + Cl–<\/sup>(aq) \u2192\u00a0AgCl(s) \u00a0 \u00a0\u0394H = -0.1 y kJ mol-1<\/sup><\/li>\n
\n<\/li>\n
\n(a) The two solutions are mixed quickly to avoid too much heat loss to the surroundings.
\n(b) A plastic cup is used because plastic is a good insulator of heat. This will reduce heat loss to the surroundings.
\n(c) The initial temperatures of the silver nitrate solution and sodium chloride solution are taken after a few minutes to let the solutions achieve a uniform temperature.
\n(d) The reacting mixture is stirred slowly throughout the activity to ensure the temperature of the mixture is uniform.
\n(e) The reading of the thermometer should be observed until the highest temperature is recorded.<\/li>\n
\n(a) When the volumes of both reactants are doubled, the number of moles of both reactants is also doubled. This makes the heat change double, 2H.
\n(b) The mass of the solution is also doubled, 2m.
\n(c) Heat change, H = mc\u03b8
\nWhen the volumes of both reactants are doubled, heat change, 2H = 2mc\u03b8.
\nCancelling the number 2 from both sides of the equation gives H = mc\u03b8.
\nThus, \u03b8\u00a0remains the same.
\nIf the sodium chloride solution is replaced with hydrochloric acid of the same concentration, the heat of precipitation is still the same. This is because both sodium chloride solution and hydrochloric acid provide the same amount of chloride ions. Sodium ions, Na+<\/sup> and hydrogen ions, H+<\/sup> are spectator ions. The reaction involves Ag+<\/sup> ions and Cl–<\/sup> ions.<\/li>\n<\/ol>\n
\nThe heat of precipitation for silver chloride is -0.1 y kJ mol-1<\/sup>.<\/p>\nHow to calculate heat of precipitation example problems with solutions<\/strong><\/h2>\n
\n[Specific heat capacity of solution: 4.2 J g-1<\/sup> \u00b0C-1<\/sup>. Density of solution: 1 g cm-3<\/sup>]
\nSolution:<\/strong>
\n
\n<\/p>\n
\nPb2+<\/sup>(aq) + SO4<\/sub>2-<\/sup>(aq) \u2192 PbSO4<\/sub>(s) \u00a0 \u00a0 \u0394H = -50.4 kJ
\nWhen 100 cm3<\/sup> of 0.5 mol dm-3<\/sup> lead(II) nitrate solution is added to 100 cm3<\/sup> of 0.5 mol dm-3<\/sup> sodium sulphate solution, lead(II) sulphate is precipitated. What is the temperature change in the reacting mixture in the experiment?
\n[Specific heat capacity of solution = 4.2 J g-1 <\/sup>\u00b0C-1<\/sup>. Density of solution = 1 g cm-3<\/sup>]
\nSolution:<\/strong>
\n<\/p>\n
\nWhat is the change in temperature if
\n(a) 100 cm3<\/sup> of 2.0 mol dm-3<\/sup>\u00a0lead(II) nitrate solution is added to 100 cm3<\/sup> of 2.0 mol dm-3<\/sup>\u00a0potassium sulphate solution?
\n(b) 50 cm3<\/sup> of 1.0 mol dm-3<\/sup>\u00a0lead(II) nitrate solution is added to 50 cm3<\/sup> of 1.0 mol dm-3<\/sup>\u00a0potassium sulphate solution?
\n(c) 50 cm3<\/sup> of 2.0 mol dm-3<\/sup>\u00a0lead(II) nitrate solution is added to 50 cm3<\/sup> of 1.0 mol dm-3<\/sup>\u00a0potassium sulphate solution?
\n(d) 50 cm3<\/sup> of 2.0 mol dm-3<\/sup>\u00a0potassium sulphate solution is replaced by 50 cm3<\/sup> of 2.0 mol dm-3<\/sup>\u00a0sodium sulphate solution?
\nSolution:
\n(a)<\/strong> The volumes of both reactants are doubled.
\nBy calculation:
\n<\/strong>When 50 cm3<\/sup> of 2.0 mol dm-3<\/sup> of lead(II) nitrate solution and 50 cm3<\/sup> of 2.0 mol dm-3<\/sup> potassium sulphate solution are used,
\n
\nWhen 100 cm3<\/sup> of 2.0 mol dm-3<\/sup> of lead(II) nitrate solution and 100 cm3<\/sup> of 2.0 mol dm-3<\/sup> potassium sulphate solution are used,
\n
\nBy deduction:
\n<\/strong>When the volumes of both reactants are doubled, the number of moles of both reactants is also doubled. This makes the heat change double, 2H.
\nThe mass of the solution is also doubled, 2m.
\n<\/p>\n
\nBy calculation:
\n<\/strong>When 50 cm3<\/sup> of 1.0 mol dm-3<\/sup> lead(II) nitrate solution and 50 cm3<\/sup> of 1.0 mol dnr3<\/sup>potassium sulphate solution are used,
\n
\nBy deduction:<\/strong>
\nWhen the concentrations of both reactants are halved, the number of moles of both reactants is also halved.
\n<\/p>\n
\n
\nPotassium sulphate is the limiting substance, thus the calculation is based on 0.05 mole of SO4<\/sub>2-<\/sup>\u00a0ions.
\nSince the number of moles of substances is halved, the heat produced is also halved.
\nMass of solution is the same, thus
\n<\/p>\n
\nThis reaction involves the reaction between lead(II) ions and sulphate ions to form lead(II) sulphate as shown below.
\nPb2+<\/sup>(aq) + SO4<\/sub>2-<\/sup>(aq) \u2192 PbSO4<\/sub>(s)
\nBoth potassium sulphate solution and sodium sulphate solution provide the same number of moles of sulphate ions.
\nThus, the temperature change is the same, \u03b8\u00a0= 10\u00b0C<\/p>\n","protected":false},"excerpt":{"rendered":"