{"id":13141,"date":"2020-11-30T05:34:06","date_gmt":"2020-11-30T00:04:06","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=13141"},"modified":"2020-11-30T17:12:58","modified_gmt":"2020-11-30T11:42:58","slug":"constructing-ionic-equations-using-continuous-variation-method","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/constructing-ionic-equations-using-continuous-variation-method\/","title":{"rendered":"Constructing ionic equations using the continuous variation method"},"content":{"rendered":"
1.<\/strong> 10 cm3<\/sup> of 0.25 mol dm-3<\/sup> lead(II) nitrate solution reacts completely with 5 cm3<\/sup> of 1.0 mol dm-3<\/sup> potassium iodide solution. A yellow precipitate of lead(II) iodide is formed. Construct the ionic equation for the formation of lead(II) iodide. People also ask<\/strong><\/p>\n Aim:<\/strong> To construct an ionic equation for the formation of lead(II) chromate(VI). Results:<\/strong><\/p>\n Graph:<\/strong> Conclusion:<\/strong> Constructing ionic equations using the continuous variation method The ionic equation for the formation of an insoluble salt can be constructed if we know the number of moles of cation and anion reacted together to form 1 mole of the insoluble salt. For example: (a) 1 mole of silver chromate(VI) is formed from 2 moles […]<\/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":[4726,4729,4728,4727,4725,4724],"yoast_head":"\n
\nSolution:
\n
\n<\/strong><\/p>\n\n
Constructing ionic equations experiment<\/strong><\/h2>\n
\nProblem statement:<\/strong> How to construct an ionic equation for the formation of lead(II) chromate(VI)?
\nHypothesis:<\/strong> As the volume of potassium chromate(VI) solution increases, the height of the yellow precipitate increases until all the lead(II) nitrate has reacted.
\nVariables:<\/strong>
\n(a) Manipulated variable : Volume of potassium chromate(VI) solution
\n(b) Responding variable : Height of yellow lead(II) chromate(VI) precipitate
\n(c) Controlled variables : Volume and concentration of lead(II) nitrate solution, concentration of potassium chromate(VI) solution, size of test tubes
\nMaterials:<\/strong> 0.5 mol dm-3<\/sup> lead(II) nitrate solution and 0.5 mol dm-3<\/sup> potassium chromate(VI) solution.
\nApparatus:<\/strong> 7 test tubes, test tube rack, burettes, metre rule, stopper, dropper, retort stand and clamp.
\nProcedure:<\/strong><\/p>\n\n
\n\n
\n Test tube<\/strong><\/td>\n 1<\/td>\n 2<\/td>\n 3<\/td>\n 4<\/td>\n 5<\/td>\n 6<\/td>\n 7<\/td>\n<\/tr>\n \n Volume of 0.5 mol dm-3<\/sup> lead(II) nitrate solution<\/strong><\/td>\n 5<\/td>\n 5<\/td>\n 5<\/td>\n 5<\/td>\n 5<\/td>\n 5<\/td>\n 5<\/td>\n<\/tr>\n \n Volume of 0.5 mol dm-3<\/sup> potassium chromate(VI) solution (cm3<\/sup>)<\/strong><\/td>\n 1<\/td>\n 2<\/td>\n 3<\/td>\n 4<\/td>\n 5<\/td>\n 6<\/td>\n 7<\/td>\n<\/tr>\n \n Height of precipitate (cm)<\/strong><\/td>\n 0.6<\/td>\n 1.2<\/td>\n 1.8<\/td>\n 2.4<\/td>\n 3.0<\/td>\n 3.0<\/td>\n 3.0<\/td>\n<\/tr>\n \n Colour of solution above precipitate<\/strong><\/td>\n Colourless<\/td>\n Colourless<\/td>\n Colourless<\/td>\n Colourless<\/td>\n Colourless<\/td>\n Yellow<\/td>\n Yellow<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
\nA graph of the height of precipitate against the volume of potassium chromate(VI) solution added is plotted.
\nInterpreting data:<\/strong>
\nFrom the graph:
\nVolume of potassium chromate(VI) solution required to completely react with 5.0 cm3<\/sup> of lead(II) nitrate solution = 5.00 cm3
\n<\/sup>
\nNumber of moles of Pb2+<\/sup> ions in 5.00 cm3<\/sup> of 0.5 mol dm-3<\/sup> lead(II) nitrate solution in each of the test tubes
\n
\nDiscussion:<\/strong><\/p>\n\n
\nThe ionic equation for the formation of lead(II) chromate(VI) can be obtained from a precipitation reaction. The hypothesis is accepted.<\/p>\n","protected":false},"excerpt":{"rendered":"