{"id":3950,"date":"2023-10-06T08:19:03","date_gmt":"2023-10-06T08:19:03","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3950"},"modified":"2023-10-06T08:19:51","modified_gmt":"2023-10-06T08:19:51","slug":"week-9-ss3-second-term-chemistry-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-9-ss3-second-term-chemistry-notes\/","title":{"rendered":"Week 9 &#8211; SS3 Second Term Chemistry Notes"},"content":{"rendered":"<p><strong>WEEK NINE<br \/>\n<\/strong><strong>TOPIC: VOLUMETRIC ANALYSIS<br \/>\n<\/strong><strong>CONTENT<br \/>\n<\/strong><\/p>\n<ul>\n<li>Calculation Based on Percentage Purity and Impurity of substances.\n<\/li>\n<li>Percentage\/amount of water of crystallization,\n<\/li>\n<li>Molar mass of the acidic base\n<\/li>\n<li>Solubility of substances\n<\/li>\n<li>Volume of gases\n<\/li>\n<li>Mole ratio of acid to base\n<\/li>\n<\/ul>\n<p>\u00a0<strong>Volumetric Analysis<br \/>\n<\/strong>Volumetric analysis involves acid base titration.<\/p>\n<p>\u00a0<strong>Mole Ratio<br \/>\n<\/strong>Mole ratio is the ratio of the reacting species.  This determines the ratio of the acid that would react with the base.<br \/>\nExamples are<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se1.png\" alt=\"\"\/>1.  \u00a0\u00a0\u00a0\u00a0H<sub>2<\/sub>SO<sub>4<\/sub>  +  2NaOH\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0      Na<sub>2<\/sub>SO<sub>4<\/sub> + 2H<sub>2<\/sub>O<\/p>\n<p>\u00a0CaVa   =  \u00bd<br \/>\nCbVb<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se2.png\" alt=\"\"\/>2.\u00a0\u00a0\u00a0\u00a02HCl +  Na<sub>2<\/sub>CO<sub>3 <\/sub>             2NaCl +H<sub>2<\/sub>O + CO<sub>2<\/sub><br \/>\n\t\t\u00a0\u00a0\u00a0\u00a0CaVa =    2<br \/>\n\u00a0\u00a0\u00a0\u00a0CbVb       1<\/p>\n<p>\u00a0<strong>EVALUATION<br \/>\n<\/strong><\/p>\n<ol>\n<li>What is volumetric analysis\n<\/li>\n<li>\n<div>Give the ratio of the reaction species in the following chemical reactions\n<\/div>\n<p><img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se3.png\" alt=\"\"\/>a.   CaCO<sub>3<\/sub> + 2 HCl \u00a0\u00a0\u00a0\u00a0    CaCl<sub>2     <\/sub>+   H<sub>2<\/sub>O    +   CO<sub>2<\/sub><br \/>\n\t\t\t\t<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se4.png\" alt=\"\"\/>b.  KHCO<sub>3<\/sub> + 2HCl\u00a0\u00a0\u00a0\u00a0    KCl +      H<sub>2<\/sub>O    +    CO<sub>2<\/sub>\n\t\t\t\t<\/li>\n<\/ol>\n<p>\u00a0<strong>Calculation Involving Titration<br \/>\n<\/strong>1.  Mole Ratio<br \/>\nA is a solution of an acid hydrogen chloride .B is a solution of sodium trioxocarbonate(iv) containing 0.05 mole per dm<sup>3<\/sup> solution A was titrated against 25cm<sup>3<\/sup> of solution B, using methyl orange as indicator during the process, the following data were obtained.<\/p>\n<p>\u00a0Burette reading (cm<sup>3<\/sup>) \u00a0\u00a0\u00a0\u00a0         \u00a0\u00a0\u00a0\u00a0Rough \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01<sup>st\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup> 2<sup>nd\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/sup> 3<sup>rd<\/sup><br \/>\n\t\tFinal burette reading (cm<sup>3<\/sup>)\u00a0\u00a0\u00a0\u00a024.65\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a048.95\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.30<br \/>\nInitial burette reading (cm<sup>3<\/sup>)\u00a0\u00a0\u00a0\u00a00.00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.65\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.00\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00.00<br \/>\nVolume of acid used (cm<sup>3<\/sup>)\u00a0\u00a0\u00a0\u00a024.65\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.30\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.30.<\/p>\n<p>\u00a0<\/p>\n<ol>\n<li>Calculate the average titre value\n<\/li>\n<li>Calculate the concentration of the acid in moldm<sup>3<\/sup>.\n<\/li>\n<li>Calculate the concentration of the acid in g\/dm<sup>3<\/sup>.\n<\/li>\n<\/ol>\n<p>The equation of the reaction<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se5.png\" alt=\"\"\/>NaCO<sub>3<\/sub> + 2HCl             \u00a0\u00a0\u00a0\u00a0    2NaCl +H<sub>2<\/sub>O + CO<sub>2<\/sub><br \/>\n\t\tSolution<br \/>\n1. Average titre value =     24.30 + 24.30 + 24.30<br \/>\n\u00a0\u00a0\u00a0\u00a03<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0          = 24.30cm<sup>3<\/sup><\/p>\n<p>\u00a02. Concentration of A in moldm<sup>3<\/sup><br \/>\n\t\t    from\u00a0\u00a0\u00a0\u00a0<\/p>\n<p>\t\t\tCaVa = Na<br \/>\n                     CbVb    Nb<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Ca x 24.30    =  2<br \/>\n                        0.05 x 25\u00a0\u00a0\u00a0\u00a0  1<br \/>\n\u00a0\u00a0\u00a0\u00a0Ca =  0.05 x 25 x 2<br \/>\n                        24.30<br \/>\n\u00a0\u00a0\u00a0\u00a0Ca = 0.103moldm<sup>3<\/sup>.<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0OR<br \/>\nFrom no  of mole  = Conc. In moldm-3  X vol\/dm<sup>3<\/sup><br \/>\n\t\tNo of moles = 0.05 x  25<br \/>\n                                  1000<br \/>\nequation of the reaction.<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se6.png\" alt=\"\"\/>Na<sub>2<\/sub>CO<sub>3<\/sub>  + 2HCl  \u00a0\u00a0\u00a0\u00a02NaCl + H<sub>2<\/sub>O + CO<sub>2<\/sub><\/p>\n<ol>\n<li>\n<div>:  2\n<\/div>\n<p>1 mole of Na<sub>2<\/sub>CO<sub>3<\/sub> react with 2 moles of HCl<br \/>\n:. 0.00125 mole of Na<sub>2<\/sub>CO<sub>3<\/sub> will require 0.00123 x 2 of HCl<br \/>\n:. No of mole of A = 0.0025 mole<br \/>\nFrom conc of A in moldm-3 =  No of mole<br \/>\n \u00a0\u00a0\u00a0\u00a0       Volume in dm3<br \/>\n\u00a0\u00a0\u00a0\u00a0=  0.0025       \u00d7  10000<br \/>\n \u00a0\u00a0\u00a0\u00a0 24.30<br \/>\n\t\t\t\t\u00a0\u00a0\u00a0\u00a01000<br \/>\n\u00a0\u00a0\u00a0\u00a0<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0  0.0025 x 1000<br \/>\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a024.30.<br \/>\n\u00a0\u00a0\u00a0\u00a0= 0.103moldm<sup>3<\/sup><br \/>\n\t\t\t\t3. Concentration of A in g\/dm<sup>3<br \/>\n<\/sup><br \/>\n\u00a0From:- conc in g\/dm3 = conc in moldm<sup>-3<\/sup> x molar mass <\/p>\n<p>\u00a0Molar mass of HCl = 1 + 35.5  = 36.5 g\/mol.<br \/>\n:. Conc in g\/dm3 = 0.103 x 36.5<br \/>\n                          = 3.76g\/dm<sup>3<\/sup><\/p>\n<p>\u00a0<strong>PERCENTAGE PURITY AND IMPURITY<br \/>\n<\/strong>During the titration process of an impure acid or base is titrated only the pure part of either acid or base react with the base or acid.  Therefore the percentage (%) purity or impurity can be calculated.<br \/>\n% purity   = Conc in g\/dm<sup>3 <\/sup>of pure solution     X     100<br \/>\n                    Conc in g\/dm<sup>3<\/sup> of impure solution         1<br \/>\n% impurity =  conc of impure \u2013 conc of pure   X  100<br \/>\n                        conc in g\/dm<sup>3<\/sup> of impure                 1<br \/>\nMass of pure substance = Conc of pure in moldm<sup>-3<\/sup> x Molar Mass<br \/>\nMass of impurity  = Conc of impure \u2013 pure<\/p>\n<p>\u00a0Example<br \/>\nA is a solution of 020mole of HCl per dm<sup>3<\/sup>. B is a solution of an impure  sodium trioxocarbonate(iv) containing 3.0g per 250cm<sup>3<\/sup>.<br \/>\na.  Calculate the<br \/>\n(i)\u00a0 \u00a0percentage purity of A<br \/>\n(ii)  percentage impurity of A<br \/>\nVa = 20.40cm<sup>3<\/sup>     Vb = 25.00cm<sup>3<\/sup><\/p>\n<p>\u00a0The equation of reaction<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se7.png\" alt=\"\"\/>Na<sub>2<\/sub>CO<sub>3<\/sub> + 2HCl  \u00a0\u00a0\u00a0\u00a02NaCl + H<sub>2<\/sub>O + CO<sub>2<\/sub><br \/>\n\t\t\t\t(Na = 23 C= 12 O = 16 H = 1, Cl = 35.5)<br \/>\nSolution<\/p>\n<p>\t\t\t\t\tCaVa  = na<br \/>\n\t\t\t\t     CbVb    nb<\/p>\n<p>\u00a0<br \/>\n\t\t\t\t\t 0.20 x 20.40   =  2<br \/>\n      25 x cb              1<\/p>\n<p>\u00a0Cb = 0.20 x 20.40 x 1<br \/>\n\t\t\t\t                 25 x 2<br \/>\nCb = 0.0823 moldm<sup>3<\/sup><br \/>\n\t\t\t\tConc in g\/dm<sup>3<\/sup> of pure<br \/>\nFrom<br \/>\nConc in g\/dm<sup>3<\/sup> = Moldm<sup>3<\/sup> x molar mass<br \/>\nMolar mass of Na<sub>2<\/sub>CO<sup>3<\/sup> = 2(23) + 12 + 3 (16)<br \/>\nMolar mass of Na<sub>2<\/sub>CO<sub>3<\/sub> = 106g\/mol<br \/>\n:. Conc in g\/dm<sup>3 <\/sup>of pure = 0.082 x 106<br \/>\n                                     = 8.692 g\/dm<sup>3<\/sup><\/p>\n<p>\u00a0Conc of impure Na<sub>2<\/sub>XO<sub>3<\/sub><br \/>\n\t\t\t\t250 cm<sup>3<\/sup> dissolve  3.0g of Na<sub>2<\/sub>CO<sub>3<\/sub><br \/>\n\t\t\t\t1 cm3 dissolves    3.0   X 1000<br \/>\n                             250<br \/>\n                           = 12.0g\/dm<sup>3<\/sup><br \/>\n\t\t\t\t1. :. % purity  =  Conc of pure X 1000<br \/>\n\t\t\t\t                          Conc of impure      1<br \/>\n                   =  8. 69     X  100<br \/>\n                          12              1<br \/>\n                   = 72.4%<\/p>\n<p>\u00a0% impurity  =  Conc of impure \u2013 pure  X        100<br \/>\n                           Conc of impure                   1<br \/>\n% impurity  =    12 \u2013 8.6g X      100<br \/>\n\t\t\t\t                             12\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 1<br \/>\n                   = 27.6%<\/p>\n<p>\u00a0<strong>PERCENTAGE AMOUNT OF WATER OF CRYSTALLIZATION<br \/>\n<\/strong>Water of crystallization in the wager given off when an hydrated salt is heated or exposed to the atmosphere<br \/>\nHydrated salt does not contain water<br \/>\nAmount of water of crystallization is calculated as follows:<\/p>\n<p>\t\t\t\t\t Conc of anhydrous\u00a0\u00a0\u00a0\u00a0  =    moalr mass of anhydrous<br \/>\n    Conc of hydrated\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0  molar mass of hydrated<\/p>\n<p>\u00a0<strong>Percentage Water of Crystallization is calculated as follows:<br \/>\n<\/strong>%  water of crystallization\u00a0\u00a0\u00a0\u00a0=    Hydrated \u2013 Anhydrous    X         100<br \/>\n                                                                    Hydrated                  1<\/p>\n<p>\u00a0Example<br \/>\nSolution A is a solution of hydrogen chloride acid containing 0.095 moldm<sub>3<\/sub> of solution.<br \/>\nB is a solution of hydrated salt Na<sub>2<\/sub>CO<sub>3<\/sub>. XH<sub>2<\/sub>O containing 3.94g which was made up to 250cm<sub>3<\/sub> of solution with distilled water<br \/>\nVa = 29.00cm<sup>3<\/sup>, Vb = 25.00cm<sup>3.<\/sup><br \/>\n\t\t\t\tCalculate the<br \/>\nI.  value of X<br \/>\nII. percentage of water of crystallization.<\/p>\n<p>\u00a0<br \/>\n\u00a0Equation of the reaction<br \/>\n<img decoding=\"async\" align=\"left\" src=\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100623_0819_Week9SS3Se8.png\" alt=\"\"\/>Na<sub>2<\/sub>CO<sub>3<\/sub>.XH<sub>2<\/sub>O  + 2HCl \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02NaCl + H<sub>2<\/sub>O + H<sub>2<\/sub>O + CO<sub>2<\/sub><br \/>\n\t\t\t\tSolution<br \/>\ni. Value of x<br \/>\nFrom<\/p>\n<p>\t\t\t\t\tCaVa      =    Na\u00a0\u00a0\u00a0\u00a0      CaVa       =     2<br \/>\n     CbVb           Nb                CbVb            1<\/p>\n<p>\u00a0<br \/>\n\t\t\t\t\t 0.095 x 29    =  2<br \/>\n    Cb x 25            1<br \/>\nCb =  0.095 x 29 x 1<br \/>\n          25 x 2.<br \/>\nCb = 0.0550moldm<sup>3<\/sup><br \/>\n\t\t\t\tConc in g\/dm3 of Na<sub>2<\/sub>CO<sub>3<\/sub> = moldm<sup>-3 <\/sup> x m.m<br \/>\nMolar mass of Na<sub>2<\/sub>CO<sub>3<\/sub> = 2 (23) + 12 + 3(16)  = 106 g\/mol<br \/>\nConc in g\/dm<sup>3<\/sup> = 0.055 x 106  = 5.83 g\/dm<sup>3<\/sup><br \/>\n\t\t\t\tConc in g\/dm<sup>3<\/sup> of hydrated:<\/p>\n<p>\t\t\t\t\tMass     X  1000<br \/>\n                       Volume        1<\/p>\n<p>\u00a0Conc in g\/dm<sup>3<\/sup>  =  3.94 x 1000<br \/>\n                                  250<br \/>\n                        = 15.8g\/dm<sup>3<\/sup><\/p>\n<p>\t\t\t\t\t Conc of anhydrous     =  molar mass of anhydrous<\/p>\n<p>\t\t\t\t\tConc of hydrated\u00a0\u00a0\u00a0\u00a0 molar mass of hydrated.<\/p>\n<p>\u00a0<br \/>\n\t\t\t\t\t 5.83     =   106<\/p>\n<p>\t\t\t\t\t15.76          106 x 18<br \/>\n(106 x 18x) 5.83  = 106 x 15.76<br \/>\n106 + 18x  =  106 x 15.76<br \/>\n                            5.83<br \/>\n106 + 18x = 286. 55<br \/>\n18x = 286.55 \u2013 106<br \/>\n18x = 180.55<br \/>\nx =  180.55<br \/>\n           18.<br \/>\nx = 10<br \/>\nThe salt is Na<sub>2<\/sub>CO<sub>3<\/sub>.10H<sub>2<\/sub>O<\/p>\n<p>\u00a0<strong>READING ASSIGNMENT<br \/>\n<\/strong>Practical Chemistry by Makanjuola pages 1-15.<br \/>\nNew School Chemistry by Osei Yaw Ababio pages 165 \u2013 183<br \/>\nPractical Chemistry for Schools and Colleges pages 100 &#8211; 170<\/p>\n<p>\u00a0<strong>GENERAL\u00a0EVALUATION<br \/>\n<\/strong><\/p>\n<ol>\n<li>What is volumetric analysis\n<\/li>\n<li>Name five  apparatus used in volumeric analysis.\n<\/li>\n<li>Define the following terms;   a. Indicator  b. Buffers  c. pH scale\n<\/li>\n<\/ol>\n<p>\u00a0<strong>WEEKEND ASSIGNMENT<br \/>\n<\/strong><\/p>\n<ol>\n<li>C + water give colourless solution (a) c is a soluble salt (b) c is partially dissolve in water (c) c is a filterate (d) c is a residue\n<\/li>\n<li>____ is the apparatus use to convert vapor into liquid during distillation. (a) conical flask\u00a0(b) distillation column   (c) lie-big condenser (d) round bottom flask\n<\/li>\n<li>X which fumes in most air can be suitably stored (a) under paraffin or naphtha (b) In a white bottle (c) inside a corked conical flask (d) inside a burette.\n<\/li>\n<li>The observation in bubbling SO<sub>2 <\/sub>into acidified KMnO<sub>4<\/sub> solution is (a) The solution turns to green   (b) the solution becomes decolourized  (c) no visible reaction (d) the solution turns steam\n<\/li>\n<li>The two substances that can give both H<sub>2<\/sub> and ZnSO<sub>4<\/sub> when added to H<sub>2<\/sub>SO<sub>4<\/sub> are: (a) Magnesium and Zinc (b) Magnesium and CuO (c) Sodium and NaOH (d) iron and copper\n<\/li>\n<\/ol>\n<p>\u00a0<strong>THEORY<br \/>\n<\/strong><\/p>\n<ol>\n<li>\n<div>State what would observe on\n<\/div>\n<ol>\n<li>mixing Zinc dust with CuSO<sub>4<\/sub> solution\n<\/li>\n<li>adding concentrated HNO<sub>3 <\/sub>to freshly prepared FeSO<sub>4<\/sub> solution\n<\/li>\n<\/ol>\n<\/li>\n<li>\n<div>A salt sample was suspected to be either Na<sub>2<\/sub>CO<sub>3<\/sub> or NaHCO<sub>3<\/sub>. A student who was required to identify it, tested a portion for solubility in water and for effects on litmus paper.\n<\/div>\n<ol>\n<li>What was the observation in each case?\n<\/li>\n<li>State the reason why the student&#8217;s procedure was unsuitable.\n<\/li>\n<li>Describe briefly how you would have identified the salt.\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n","protected":false},"excerpt":{"rendered":"<p>WEEK NINE TOPIC: VOLUMETRIC ANALYSIS CONTENT Calculation Based on Percentage Purity and Impurity of substances&#8230;.<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,307],"tags":[],"class_list":["post-3950","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss3-chemistry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3950","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3950"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3950\/revisions"}],"predecessor-version":[{"id":3951,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3950\/revisions\/3951"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3950"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3950"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3950"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}