{"id":1434,"date":"2024-08-17T11:30:50","date_gmt":"2024-08-17T11:30:50","guid":{"rendered":"https:\/\/www.meniit.com\/study-material\/?p=1434"},"modified":"2024-08-21T08:36:10","modified_gmt":"2024-08-21T08:36:10","slug":"stoichiometry","status":"publish","type":"post","link":"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/stoichiometry","title":{"rendered":"Stoichiometry"},"content":{"rendered":"<h2 style=\"text-align: justify;\">SECTION 1 : CONCEPT OF EQUIVALENT MASS<\/h2>\n<ul style=\"text-align: justify;\">\n<li><strong>CLASSICAL DEFINITION<\/strong><\/li>\n<\/ul>\n<p style=\"text-align: justify;\">It is defined as the number of parts by weight of a substance that <strong>combines<\/strong> with or <strong>displaces<\/strong> directly or indirectly <strong>1.008 parts<\/strong> by weight of <strong>hydrogen<\/strong> or <strong>8.0 parts<\/strong> of <strong>oxygen<\/strong> or <strong>35.5 parts<\/strong> of <strong>chlorine<\/strong>. Atomic mass and molecular mass of elements and compounds are always constant but equivalent mass may change. A compound may have different equivalent mass in different reactions.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-full wp-image-1435\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/redox-processes.png\" alt=\"redox processes\" width=\"761\" height=\"818\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/redox-processes.png 761w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/redox-processes-279x300.png 279w\" sizes=\"auto, (max-width: 761px) 100vw, 761px\" \/><\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_69_1 counter-hierarchy ez-toc-counter ez-toc-light-blue ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 ' ><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/stoichiometry\/#SECTION-2-n-FACTOR-CALCULATION\" title=\"SECTION 2: n-FACTOR CALCULATION\">SECTION 2: n-FACTOR CALCULATION<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/stoichiometry\/#SECTION-3-EXPRESSING-CONCENTRATION-OF-SOLUTIONS\" title=\"SECTION 3 : EXPRESSING CONCENTRATION OF SOLUTIONS\">SECTION 3 : EXPRESSING CONCENTRATION OF SOLUTIONS<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/stoichiometry\/#SECTION-4-TITRATIONS\" title=\"SECTION 4 : TITRATIONS\">SECTION 4 : TITRATIONS<\/a><\/li><\/ul><\/nav><\/div>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"SECTION-2-n-FACTOR-CALCULATION\"><\/span>SECTION 2: n-FACTOR CALCULATION<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><strong> Case-I : Acids &amp; Bases :<\/strong> Acids are the species which furnish H<sup>+<\/sup> ions when dissolved in a solvent.<br \/>\nFor acids, n factor is defined as the number of H<sup>+<\/sup> ions replaced by 1 mole of acid in a reaction. Note that then factor for acid is not equal to its basicity; i.e., the number of moles of replaceable H+ atoms present in one mole of acid.<\/p>\n<p style=\"text-align: justify;\">Bases are the species, which furnish OH<sup>\u2013<\/sup> ions when dissolved in a solvent. For bases, <strong>n-factor is defined as the number defined as the number<\/strong> of OH<sup>\u2013<\/sup> ions replaced by 1 mole of base in a reaction. Note that n factor is not equal to its acidity, i.e., the number of moles of replaceable OH\u2013 ions present in 1 mole of base.<br \/>\nFor instance,<\/p>\n<table style=\"height: 560px;\" width=\"704\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for HCl<\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of NaOH<\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for HNO<sub>3<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of NH<sub>4<\/sub>OH<\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for H<sub>2<\/sub>SO<sub>4<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of Zn(OH)<sub>2<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for H<sub>3<\/sub>PO<sub>4<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2 or 3<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of Ca(OH)<sub>2<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for H<sub>3<\/sub>PO<sub>3<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of Al(OH)<sub>3<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2 or 3<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for H<sub>3<\/sub>PO<sub>2<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of Ba(OH)<sub>2<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1 or 2<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center;\" width=\"114\">n-factor for H<sub>3<\/sub>BO<sub>3<\/sub><\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<td style=\"text-align: center;\" width=\"119\">n-factor of CsOH<\/td>\n<td style=\"text-align: center;\" width=\"62\">1<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\"><strong>Case-II :<\/strong> If the salt involved in a reaction does not undergo oxidation state then the n factor for such<br \/>\nsalts is defined as the total cationic or total anionic charge present in 1 mole of the salt or simply the LCM of<br \/>\nthe charges on the cation and anion. For instance in the reaction, Na<sub>3<\/sub>PO<sub>4<\/sub> + BaCl<sub>2<\/sub> NaCl + Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub>.<br \/>\nFor instance, in Na<sub>3<\/sub>PO<sub>4<\/sub> the LCM of charge on Na+ and is 3 and PO<sub>4<\/sub><sup>3\u2013<\/sup> so is the total positive charge<br \/>\non 3Na+ ions.<br \/>\nSame way, if we wish to find the n-factor for Ba<sub>3<\/sub>(PO<sub>4<\/sub>)<sub>2<\/sub> then, it will be given as the LCM of charges on<br \/>\nBa<sup>2+<\/sup> and PO<sub>4<\/sub><sup>3\u2013<\/sup> that is 6.<\/p>\n<p style=\"text-align: justify;\"><strong>Case-III : n-Factor in Redox Reactions :<\/strong> If an element in a compound undergoes a change in oxidation state then we must look at the change in oxidation state and use the under-mentioned formula for all redox changes :<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1437 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant.png\" alt=\"oxidant\" width=\"684\" height=\"85\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant.png 684w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant-300x37.png 300w\" sizes=\"auto, (max-width: 684px) 100vw, 684px\" \/><\/p>\n<p style=\"text-align: justify;\">For example, let us calculate the n factor KMnO4 for the given chemical change.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1438 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant1.png\" alt=\"oxidant1\" width=\"251\" height=\"45\" \/><\/p>\n<p style=\"text-align: justify;\">In this reaction, oxidation state of Mn changes from +7 to +2. Thus, KMnO<sub>4<\/sub> is acting as oxidising agent, since it is reduced.<\/p>\n<p style=\"text-align: justify;\"><strong>\u2234\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 n-factor<\/strong> of KMnO<sub>4<\/sub> = 5 in acidic media<\/p>\n<p style=\"text-align: justify;\">Similarly,<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1439 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant2.png\" alt=\"oxidant2\" width=\"821\" height=\"80\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant2.png 821w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant2-300x29.png 300w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/oxidant2-768x75.png 768w\" sizes=\"auto, (max-width: 821px) 100vw, 821px\" \/><\/p>\n<p style=\"text-align: justify;\">It can be seen that in all the above chemical changes, KMnO<sub>4<\/sub> is acting as oxidising agent, yet its <strong>n-factor<\/strong> is not same in all reactions. Thus, the <strong>n-factor<\/strong> of a compound is not fixed, it depends on the type and the extent of reaction it undergoes.<\/p>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"SECTION-3-EXPRESSING-CONCENTRATION-OF-SOLUTIONS\"><\/span>SECTION 3 : EXPRESSING CONCENTRATION OF SOLUTIONS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><strong> Solution<\/strong> is a homogenous mixture of two or more components in which intermingling particles are of atomic or molecular dimensions. A solution consists of a dissolved substance known as <strong>solute<\/strong> and the substance in which the solute is dissolved is known as <strong>solvent<\/strong>. The <strong>concentration<\/strong> of a solute means the quantity of solute dissolved per unit volume of solution, or per unit quantity of solvent.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1440 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/solute.png\" alt=\"solute\" width=\"490\" height=\"66\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/solute.png 490w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/solute-300x40.png 300w\" sizes=\"auto, (max-width: 490px) 100vw, 490px\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>NOTE<\/strong> : While discussing various methods for expressing concentration, we have taken solute as <strong>B<\/strong> dissolved in solvent <strong>A<\/strong> and W<sub>B<\/sub> as grams of solute and W<sub>A<\/sub> as grams of solvent.<\/p>\n<p style=\"text-align: justify;\"><strong> Molarity (M)<\/strong> is expressed as moles of solute contained in one litre of solution or it is also taken as millimoles of solute in 1000 cc (ml) of solution. It is denoted by <strong>M.<\/strong><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1441 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/molarity.png\" alt=\"molarity\" width=\"441\" height=\"107\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/molarity.png 441w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/molarity-300x73.png 300w\" sizes=\"auto, (max-width: 441px) 100vw, 441px\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Molarity (M)<\/strong> is expressed as number of moles of solute dissolved in 1000 gms (1 Kg) of solvent. It is<br \/>\ndenoted by <strong>m.<\/strong><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1442 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/molaritym.png\" alt=\"molaritym\" width=\"147\" height=\"65\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Normality (N)<\/strong> is expressed as the number of gram equivalents (gm eq) of solute contained in one litre of solution or it can also be taken as number of milliequivalents (meq) in 1000 cc (ml) of solution. It is denoted by <strong>N.<\/strong><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1443 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/normality-of-solution.png\" alt=\"normality of solution\" width=\"600\" height=\"111\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/normality-of-solution.png 600w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/normality-of-solution-300x56.png 300w\" sizes=\"auto, (max-width: 600px) 100vw, 600px\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Strength of a solution<\/strong> is generally expressed as grams of solute in one litre of solution.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1444 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/strength.png\" alt=\"strength\" width=\"482\" height=\"80\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/strength.png 482w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/strength-300x50.png 300w\" sizes=\"auto, (max-width: 482px) 100vw, 482px\" \/><\/p>\n<p style=\"text-align: justify;\"><strong> Diluting a Solution: <\/strong>Whenever a given solution of known concentration, i.e., normality and molarity (known as standard solution) is diluted by adding more of solvent, the number of millimoles (or millequivalents) of solutes remain unchanged. The concentration of solution however changes.<\/p>\n<p style=\"text-align: center;\">In such cases,\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 M<sub>1<\/sub>V<sub>1<\/sub>\u00a0= M<sub>2<\/sub>V<sub>2<\/sub>\u00a0(M<sub>1<\/sub>\u00a0&amp; V<sub>1<\/sub>a re molarity and volume of original solution<\/p>\n<p style=\"text-align: center;\">\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(M<sub>2<\/sub> and V<sub>2<\/sub> are molarity and volume of diluted solution)<\/p>\n<p style=\"text-align: justify;\"><strong> Mass fraction\u00a0<\/strong>is the fractional part of a component that is contributed by it to the total mass of solution.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1445 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/mass-fraction.png\" alt=\"mass fraction\" width=\"325\" height=\"130\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/mass-fraction.png 325w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/mass-fraction-300x120.png 300w\" sizes=\"auto, (max-width: 325px) 100vw, 325px\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Mole fraction\u00a0<\/strong>is the fractional part of the moles that is contributed by each component to the total number of moles that comprises the solution. In containing n<sub>A<\/sub> moles of solvent and n<sub>B<\/sub> moles of solute :<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1446 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/mole-fraction.png\" alt=\"mole fraction\" width=\"300\" height=\"118\" \/><\/p>\n<h4 style=\"text-align: justify;\">Volume Fraction<\/h4>\n<p style=\"text-align: justify;\">It is <strong>volume of solute<\/strong> over the <strong>total volume of solution<\/strong>.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1447 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/volume-fraction.png\" alt=\"volume fraction\" width=\"348\" height=\"58\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/volume-fraction.png 348w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/volume-fraction-300x50.png 300w\" sizes=\"auto, (max-width: 348px) 100vw, 348px\" \/><\/p>\n<p style=\"text-align: justify;\">(Remember that fractions of all types always add up to give unity)<\/p>\n<h4 style=\"text-align: justify;\">Parts per Million (ppm) and Parts per Billion (ppb)<\/h4>\n<p style=\"text-align: justify;\">This concentration term defines the<strong> number of milligrams of solute present per kilogram of solvent<\/strong> for the case of parts per million. Usually it is employed in those solutions where the solute with respect to the solution is very less.<\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1448 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/ppm-ppb.png\" alt=\"ppm ppb\" width=\"585\" height=\"112\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/ppm-ppb.png 585w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/ppm-ppb-300x57.png 300w\" sizes=\"auto, (max-width: 585px) 100vw, 585px\" \/><\/p>\n<h3 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"SECTION-4-TITRATIONS\"><\/span>SECTION 4 : TITRATIONS<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p style=\"text-align: justify;\"><strong>Titration<\/strong> is an experiment that is done to determine the unknown concentration of a substance. The substance whose concentration is known is called the <strong>Titrant or titrator<\/strong>. A standard solution of the titrant is prepared and added to the burette. The <strong>Titrand<\/strong> or <strong>Analyte<\/strong> is the substance that is measured in a fixed volume with the help of a pipette and placed in a conical flask (see diagram below).<\/p>\n<p style=\"text-align: justify;\">As expressed in the diagram below, H<sub>2<\/sub>SO<sub>4<\/sub> of unknown concentration is placed in a conical flask as an analyte. Few drops of <strong>phenolphthalein<\/strong> indicator as added to it. The solution will remain to be colourless as phenolphthalein will show pink <strong>colour in base media<\/strong> (pH 8.0 to 10.0).<\/p>\n<p style=\"text-align: justify;\">NaOH of known concentration is placed in the burette and slowly added to H<sub>2<\/sub>SO<sub>4<\/sub> till a faint pink colour is noted. This is the <strong>end point of the titration<\/strong>e\u00a0at which <strong>gram equivalents of both NaOH and H<sub>2<\/sub>SO<sub>4<\/sub> are equal.<\/strong><\/p>\n<p style=\"text-align: justify;\"><img loading=\"lazy\" decoding=\"async\" class=\"size-full wp-image-1449 aligncenter\" src=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/titration.png\" alt=\"titration\" width=\"726\" height=\"249\" srcset=\"https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/titration.png 726w, https:\/\/www.meniit.com\/study-material\/wp-content\/uploads\/2024\/08\/titration-300x103.png 300w\" sizes=\"auto, (max-width: 726px) 100vw, 726px\" \/><\/p>\n<p style=\"text-align: justify;\">Titrations can belong to the following categories :<\/p>\n<p style=\"text-align: justify;\"><strong> Simple Titrations :<\/strong><\/p>\n<p style=\"text-align: justify;\">A known volume of the solution of unknown concentration is taken in a flask and required reagents are added to it. The solution of known concentration is added from the burette in the solution of unknown concentration till the latter reacts completely. This process is called titration. At the end point (equivalence point) the equivalents or milliequivalents of the two reacting substances are equal.<\/p>\n<p style=\"text-align: center;\">Volume of solution (A) = V<sub>A<\/sub> litres<\/p>\n<p style=\"text-align: center;\">Normality of solution (A) = N<sub>A<\/sub><\/p>\n<p style=\"text-align: center;\">Equivalents of substance (A) = N<sub>A<\/sub>V<sub>A<\/sub><\/p>\n<p style=\"text-align: center;\">Similarly, equivalents of substance (B) = N<sub>B<\/sub>V<sub>B<\/sub><\/p>\n<p style=\"text-align: center;\">At the equivalence point (end point) the equivalents (not the moles) of the two substance are equal.<\/p>\n<p style=\"text-align: center;\">N<sub>A<\/sub>V<sub>A<\/sub> (litre) = N<sub>B<\/sub> x V<sub>B<\/sub> (litre)<\/p>\n<p style=\"text-align: justify;\"><strong>Redox Titrations<\/strong><\/p>\n<p style=\"text-align: justify;\">In a redox titration, an oxidant is estimated by adding reductant or vice versa.<\/p>\n<p style=\"text-align: justify;\">For example, Fe<sup>2+<\/sup> ions can be estimated by titration against acidified KMnO4 solution when Fe<sup>2+<\/sup> ions are oxidised to Fe<sup>3+<\/sup> ions and KMnO<sub>4<\/sub> is reduced to Mn<sup>2+<\/sup> in the presence of acidic medium. KMnO<sub>4<\/sub> functions as self indicator as its purple colour is discharged at the equivalence point.<\/p>\n<p style=\"text-align: center;\">MnO<sup>\u2013<\/sup><sub>4<\/sub> + 8H<sup>+<\/sup> + 5e<sup>\u2013<\/sup> <b>\u2192<\/b> Mn<sup>2+<\/sup> + 4H2O<\/p>\n<p style=\"text-align: center;\">Fe<sup>2+<\/sup> <b>\u2192<\/b> Fe<sup>3+<\/sup> + e<sup>\u2013<\/sup> ] 5<\/p>\n<p style=\"text-align: center;\">MnO<sup>\u2013<\/sup><sub>4<\/sub> (purple) + 8H<sup>+<\/sup> + 5Fe<sup>2+<\/sup> <b>\u2192<\/b> Mn<sup>2+<\/sup> + 5Fe<sup>3+<\/sup> + 4H<sub>2<\/sub>O<\/p>\n<p style=\"text-align: center;\">(n = 5)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (n = 1)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 Colourless.<\/p>\n<div class=\"newspaper-x-tags\"><strong>TAGS: <\/strong><span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/concentration-of-solution\" rel=\"tag\">concentration of solution<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/equivalent-mass\" rel=\"tag\">equivalent mass<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/mass-fraction\" rel=\"tag\">Mass fraction<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/molality\" rel=\"tag\">Molality<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/molarity\" rel=\"tag\">Molarity<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/mole-fraction\" rel=\"tag\">Mole fraction<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/normality\" rel=\"tag\">Normality<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/redox-titration\" rel=\"tag\">redox titration<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/stoichiometry\" rel=\"tag\">stoichiometry<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/titration\" rel=\"tag\">Titration<\/a><\/span><a href=\"https:\/\/www.meniit.com\/study-material\/tag\/volume-fraction\" rel=\"tag\">Volume Fraction<\/a> <\/div>\n","protected":false},"excerpt":{"rendered":"<p>SECTION 1 : CONCEPT OF EQUIVALENT MASS CLASSICAL DEFINITION It is defined as the number of parts by weight of&nbsp;&nbsp;&#8230;.<a class=\"read_more\" href=\"https:\/\/www.meniit.com\/study-material\/neet\/class-11th\/chemistry\/stoichiometry\" rel=\"nofollow\">Read More >><\/a><\/p>\n","protected":false},"author":6,"featured_media":1484,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"om_disable_all_campaigns":false,"rank_math_lock_modified_date":false,"footnotes":""},"categories":[268,241,240],"tags":[478,476,482,480,479,483,481,486,477,485,484],"class_list":["post-1434","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-chemistry","category-class-11th","category-neet","tag-concentration-of-solution","tag-equivalent-mass","tag-mass-fraction","tag-molality","tag-molarity","tag-mole-fraction","tag-normality","tag-redox-titration","tag-stoichiometry","tag-titration","tag-volume-fraction"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts\/1434","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/comments?post=1434"}],"version-history":[{"count":4,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts\/1434\/revisions"}],"predecessor-version":[{"id":1476,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/posts\/1434\/revisions\/1476"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/media\/1484"}],"wp:attachment":[{"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/media?parent=1434"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/categories?post=1434"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.meniit.com\/study-material\/wp-json\/wp\/v2\/tags?post=1434"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}