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Concept 05 / 06Some Basic Concepts of Chemistry

Equivalent Weight, Normality and Volumetric Analysis

The n-factor for acids, bases, oxidising and reducing agents and salts, equivalent weight E = M/n, gram equivalents and normality N = n-factor x M, the law of equivalence in titrations, back titration and double titration.

The equivalent concept simplifies titration and reaction calculations: in every reaction, the reactants react in equal numbers of gram equivalents, regardless of coefficients. It requires knowing the n-factor of each substance in the particular reaction. n-factor (valence factor). Acids: number of replaceable H+\mathrm{H^{+}} ions per molecule in the reaction (basicity). HCl: 11; H2SO4\mathrm{H_{2}SO_{4}}: 22; H3PO3\mathrm{H_{3}PO_{3}} (phosphorous acid): 22, because only two H are acidic (the third is bonded to P); H3PO2\mathrm{H_{3}PO_{2}}: 11. Bases: number of OH−\mathrm{OH^{-}} ions replaced (acidity). NaOH: 11; Ca(OH)2\mathrm{Ca(OH)_{2}}: 22. Oxidising and reducing agents: the number of electrons gained or lost per formula unit. KMnO4\mathrm{KMnO_{4}} in acid (Mn7+→Mn2+\mathrm{Mn^{7+}} \rightarrow \mathrm{Mn^{2+}}): 55; in neutral or weakly basic medium (→MnO2\rightarrow \mathrm{MnO_{2}}): 33; in strongly basic medium (→MnO42−\rightarrow \mathrm{MnO_{4}^{2-}}): 11. K2Cr2O7\mathrm{K_{2}Cr_{2}O_{7}} in acid (2Cr6+→2Cr3+2\mathrm{Cr^{6+}} \rightarrow 2\mathrm{Cr^{3+}}): 66. FeSO4\mathrm{FeSO_{4}} (Fe2+→Fe3+\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}}): 11. Oxalate C2O42−→2CO2\mathrm{C_{2}O_{4}^{2-}} \rightarrow 2\mathrm{CO_{2}}: 22. Salts in non-redox reactions: total positive (or negative) charge, e.g. Na2CO3\mathrm{Na_{2}CO_{3}} reacting completely with acid: 22. The n-factor depends on the reaction, not just the substance. Na2CO3\mathrm{Na_{2}CO_{3}} has n-factor 11 when titrated only to NaHCO3\mathrm{NaHCO_{3}} (phenolphthalein end point). Equivalent weight: E=Mn-factorE = \frac{M}{n\text{-factor}}. Gram equivalents =massE=n-factor×moles= \frac{\text{mass}}{E} = n\text{-factor} \times \text{moles}. Normality: N=gram equivalentsvolume in L=n-factor×MN = \frac{\text{gram equivalents}}{\text{volume in L}} = n\text{-factor} \times M (molarity). Law of equivalence. In any reaction, gram equivalents of each reactant are equal. For titrations: N1V1=N2V2N_{1}V_{1} = N_{2}V_{2}. This holds even when the balanced equation has unequal coefficients, because the n-factors absorb the stoichiometry. For example, in the titration of oxalic acid with KMnO4\mathrm{KMnO_{4}} in acid, 2MnO4−+5C2O42−+16H+→…2\mathrm{MnO_{4}^{-}} + 5\mathrm{C_{2}O_{4}^{2-}} + 16\mathrm{H^{+}} \rightarrow \ldots: equivalents of KMnO4\mathrm{KMnO_{4}} (5×5 \times moles) equal equivalents of oxalate (2×2 \times moles), which is exactly the 2:52 : 5 mole ratio. Back titration. When a substance reacts too slowly or is insoluble, add a known excess of a reagent, then titrate the unreacted excess. Equivalents of substance == equivalents of reagent added −- equivalents of excess found. Double titration (mixtures of NaOH\mathrm{NaOH}, Na2CO3\mathrm{Na_{2}CO_{3}} and NaHCO3\mathrm{NaHCO_{3}}). With phenolphthalein, titration stops when NaOH\mathrm{NaOH} is fully neutralised and Na2CO3\mathrm{Na_{2}CO_{3}} has become NaHCO3\mathrm{NaHCO_{3}}. Methyl orange then detects the conversion of all HCO3−\mathrm{HCO_{3}^{-}} to CO2\mathrm{CO_{2}}. From the two readings, the amounts of each component follow. Key Takeaways: - n-factor: acids by replaceable H+\mathrm{H^{+}}, bases by OH−\mathrm{OH^{-}}, redox agents by electrons per formula unit. - KMnO4\mathrm{KMnO_{4}}: 55 (acid), 33 (neutral), 11 (strong base); K2Cr2O7\mathrm{K_{2}Cr_{2}O_{7}}: 66. - E=MnE = \frac{M}{n}; equivalents =n×= n \times moles; N=n×MN = n \times M. - N1V1=N2V2N_{1}V_{1} = N_{2}V_{2} at the end point. - H3PO3\mathrm{H_{3}PO_{3}} is dibasic; H3PO2\mathrm{H_{3}PO_{2}} is monobasic. - Pitfall: using a fixed n-factor for Na2CO3\mathrm{Na_{2}CO_{3}} regardless of the indicator. Brain-Teaser 1: What volume of 0.10.1 M KMnO4\mathrm{KMnO_{4}} is needed to oxidise 2525 mL of 0.20.2 M FeSO4\mathrm{FeSO_{4}} in acidic medium? Solution: Equivalents of Fe2+=0.2×1×0.025=0.005\mathrm{Fe^{2+}} = 0.2 \times 1 \times 0.025 = 0.005. Normality of KMnO4=5×0.1=0.5\mathrm{KMnO_{4}} = 5 \times 0.1 = 0.5 N. Volume =0.0050.5=0.01= \frac{0.005}{0.5} = 0.01 L =10= 10 mL. Check with moles: MnO4−+5Fe2+\mathrm{MnO_{4}^{-}} + 5\mathrm{Fe^{2+}}, so 0.0010.001 mol MnO4−\mathrm{MnO_{4}^{-}} for 0.0050.005 mol Fe2+\mathrm{Fe^{2+}}; 0.0010.1=10\frac{0.001}{0.1} = 10 mL. Both routes agree. Brain-Teaser 2: Find the equivalent weight of H3PO4\mathrm{H_{3}PO_{4}} (M=98M = 98) when it reacts with NaOH to form Na2HPO4\mathrm{Na_{2}HPO_{4}}. Solution: Two H are replaced, so the n-factor is 22 and E=49E = 49. If the product were NaH2PO4\mathrm{NaH_{2}PO_{4}} it would be 9898, and for Na3PO4\mathrm{Na_{3}PO_{4}} it would be 32.732.7. The equivalent weight is defined by the reaction. Brain-Teaser 3: 1.251.25 g of impure CaCO3\mathrm{CaCO_{3}} is dissolved in 5050 mL of 0.50.5 N HCl. The excess acid needs 2525 mL of 0.20.2 N NaOH. Find the percentage purity. Solution: HCl added =0.025= 0.025 eq; excess =0.005= 0.005 eq; so 0.0200.020 eq reacted with CaCO3\mathrm{CaCO_{3}}. The n-factor of CaCO3\mathrm{CaCO_{3}} is 22, so E=50E = 50 and the mass of pure CaCO3=0.020×50=1.0\mathrm{CaCO_{3}} = 0.020 \times 50 = 1.0 g, i.e. 1.01.25×100=80%\frac{1.0}{1.25} \times 100 = 80\% pure. Back titration measures the unreacted excess, and the difference gives the substance's equivalents.