Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Quantitative and Qualitative Analysis
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Quantitative and Qualitative Analysis
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Subject: Chemistry
Class: Senior Secondary 3
Term: 1st Term
Week: 1
Theme: Chemical World
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titrate accuratelyand performrelevant calculations analyse differentsalts for anions and cations carry put simpletest for Commongases (Oxygen,Hydrogen, CO2,etc) carry:out simpleredox titrations test for fats, oils,proteins,: starch,etc
Molar mass $\times$ Volume (in dm3) Mass of pure $Na_2CO_3$ = $0.045 M \times 106 g/mol \times 1.0 dm^3 = 4.77 g$ Percentage purity = $\frac{\text{Mass of pure } Na_2CO_3}{\text{Mass of impure sample}} \times 100\%$ Percentage purity = $\frac{4.77 g}{10.0 g} \times 100\% = 47.7\%$ Water of Crystallization: This involves determining 'x' in a hydrated salt formula (e.g., $Na_2CO_3.xH_2O$). Worked Example 3 (Water of Crystallization): 2.86 g of $Na_2CO_3.xH_2O$ was dissolved to make 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of 0.1 M HCl for complete neutralization. Calculate the value of x. (Na=23, C=12, O=16, H=1).
Solution: Equation: $Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)$ From the equation: $n_{Na_2CO_3} = 1$, $n_{HCl} = 2$ Given for HCl: $C_{HCl} = 0.1 M$, $V_{HCl} = 20.0 cm^3$ Given for $Na_2CO_3$: $V_{Na_2CO_3} = 25.0 cm^3$, $C_{Na_2CO_3} = ?$ Using $\frac{C_{Na_2CO_3} V_{Na_2CO_3}}{C_{HCl} V_{HCl}} = \frac{n_{Na_2CO_3}}{n_{HCl}}$: $\frac{C_{Na_2CO_3} \times 25.0}{0.1 \times 20.0} = \frac{1}{2}$ $50.0 C_{Na_2CO_3} = 2.0$ $C_{Na_2CO_3} = \frac{2.0}{50.0} = 0.04 M$ Mass of hydrated salt in 250 cm3 = 2.86 g Molar mass of $Na_2CO_3.xH_2O$ = Mass / Moles Moles of $Na_2CO_3$ in 250 cm3 = Molarity $\times$ Volume (in $dm^3$) = $0.04 M \times \frac{250}{1000} dm^3 = 0.04 \times 0.25 = 0.01$ moles. Since 1 mole of $Na_2CO_3.xH_2O$ contains 1 mole of $Na_2CO_3$, then moles of $Na_2CO_3.xH_2O$ = 0.01 moles. Molar mass of $Na_2CO_3.xH_2O = \frac{2.86 g}{0.01 mol} = 286 g/mol$ Molar mass of $Na_2CO_3 = 106 g/mol$ Molar mass of $xH_2O = x \times (2 \times 1 + 16) = 18x$ So, $106 + 18x = 286$ $18x = 286 - 106$ $18x = 180$ $x = \frac{180}{18} = 10$ Therefore, the formula is $Na_2CO_3.10H_2O$.
Heat of Neutralization: The heat change when one mole of water is formed from the reaction between an acid and a base. Heat change (Q) = $mc\Delta T$ Where: $m$ = mass of solution (assuming density of solution is 1 $g/cm^3$, so mass = volume in cm3) $c$ = specific heat capacity of water (4.2 J/g/K or J/g/$^\circ$C) $\Delta T$ = change in temperature ($T_{final} - T_{initial}$) Heat of Neutralization ($\Delta H_{neut}$) = $\frac{-Q}{\text{Moles of water formed}}$ (The negative sign indicates exothermic reaction).
Worked Example 4 (Heat of Neutralization): 25.0 cm3 of 1.0 M HCl at 25.0 $^\circ$C was mixed with 25.0 cm3 of 1.0 M NaOH at 25.0 $^\circ$C in a plastic cup. The highest temperature recorded was 31.8 $^\circ$C. Calculate the heat of neutralization. (Assume specific heat capacity of solution = 4.2 J/g/$^\circ$C, density of solution = 1 $g/cm^3$).
Solution: Total volume of solution = $25.0 cm^3 + 25.0 cm^3 = 50.0 cm^3$ Mass of solution (m) = $50.0 cm^3 \times 1 g/cm^3 = 50.0 g$ $\Delta T = 31.8^\circ C - 25.0^\circ C = 6.8^\circ C$ $c = 4.2 J/g/^\circ C$ $Q = mc\Delta T = 50.0 g \times 4.2 J/g/^\circ C \times 6.8^\circ C = 1428 J = 1.428 kJ$ Moles of HCl used = $C_A V_A = 1.0 M \times \frac{25.0}{1000} dm^3 = 0.025 \text{ moles}$ Moles of NaOH used = $C_B V_B = 1.0 M \times \frac{25.0}{1000} dm^3 = 0.025 \text{ moles}$ From $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$, 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of $H_2O$. So, moles of water formed = 0.025 moles. Heat of Neutralization = $\frac{-Q}{\text{Moles of water formed}} = \frac{-1.428 kJ}{0.025 \text{ mol}} = -57.12 kJ/mol$ Titration is a quantitative chemical method used to determine the concentration of an unknown solution (analyte) by reacting it with a solution of known concentration (standard solution).
Key Terms: Analyte/Titrand: The solution of unknown concentration in the conical flask.
Titrant: The solution of known concentration, added from the burette.
Standard Solution: A solution whose concentration is accurately known.
Primary Standard: A substance of high purity, stable, non-hygroscopic, and soluble, which can be weighed directly to prepare a standard solution (e.g., anhydrous Na2CO3, oxalic acid).
Secondary Standard: A solution whose concentration is determined by titration against a primary standard (e.g., NaOH, HCl).
Equivalence Point: The point in a titration where the reactants have reacted completely in stoichiometric proportions.
End Point: The point at which the indicator changes colour, signaling the completion of the reaction. Ideally, the end point should coincide with the equivalence point.
Indicator: A substance that changes colour over a narrow pH range to signal the end point of a titration.
Methyl Orange: Red in acidic medium (pH 4.4). Suitable for strong acid-strong base and strong acid-weak base titrations.
Phenolphthalein: Colourless in acidic medium (pH 10). Suitable for strong acid-strong base and weak acid-strong base titrations.
Litmus: Red in acid, blue in alkali. Less precise for titration.
Concordant Titres: Titre values that are very close to each other (e.g., within 0.20 cm3).
Calculations in Titration: The general formula for titration calculations is: $\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$ Where: $C_A$ = Concentration of acid ($mol/dm^3$) $V_A$ = Volume of acid ($cm^3$ or $dm^3$) $n_A$ = Number of moles of acid from the balanced equation $C_B$ = Concentration of base ($mol/dm^3$) $V_B$ = Volume of base ($cm^3$ or $dm^3$) $n_B$ = Number of moles of base from the balanced equation Worked Example 1 (Molarity/Concentration Calculation): In a titration, 25.0 cm3 of 0.1 M NaOH solution required 20.0 cm3 of HCl solution for complete neutralization. Calculate the concentration of the HCl solution.
Solution: Balanced chemical equation: $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(l)$ From the equation: $n_A = 1$, $n_B = 1$ Given: $C_B = 0.1 M$ $V_B = 25.0 cm^3$ $V_A = 20.0 cm^3$ $C_A = ?$ Using $\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$: $\frac{C_A \times 20.0}{0.1 \times 25.0} = \frac{1}{1}$ $20.0 C_A = 0.1 \times 25.0$ $20.0 C_A = 2.5$ $C_A = \frac{2.5}{20.0} = 0.125 M$ The concentration of the HCl solution is 0.125
M. Percentage Purity: If an impure sample of a substance reacts, the calculated mass from titration represents the mass of the pure component.
Percentage purity is then: Percentage Purity = $\frac{\text{Mass of pure substance}}{\text{Mass of impure sample}} \times 100\%$ Worked Example 2 (Percentage Purity): 10.0 g of impure sodium carbonate was dissolved in water to make 1.0 dm3 of solution. 25.0 cm3 of this solution required 22.5 cm3 of 0.1 M HCl for complete reaction. Calculate the percentage purity of the sodium carbonate. (Na=23, C=12, O=16).
Solution: Equation: $Na_2CO_3(aq) + 2HCl(aq) \rightarrow 2NaCl(aq) + H_2O(l) + CO_2(g)$ From the equation: $n_{Na_2CO_3} = 1$, $n_{HCl} = 2$ Given for HCl: $C_{HCl} = 0.1 M$, $V_{HCl} = 22.5 cm^3$ Given for $Na_2CO_3$: $V_{Na_2CO_3} = 25.0 cm^3$, $C_{Na_2CO_3} = ?$ Using $\frac{C_{Na_2CO_3} V_{Na_2CO_3}}{C_{HCl} V_{HCl}} = \frac{n_{Na_2CO_3}}{n_{HCl}}$: $\frac{C_{Na_2CO_3} \times 25.0}{0.1 \times 22.5} = \frac{1}{2}$ $2 \times C_{Na_2CO_3} \times 25.0 = 0.1 \times 22.5 \times 1$ $50.0 C_{Na_2CO_3} = 2.25$ $C_{Na_2CO_3} = \frac{2.25}{50.0} = 0.045 M$ Molar mass of $Na_2CO_3 = (2 \times 23) + 12 + (3 \times 16) = 46 + 12 + 48 = 106 g/mol$ Mass of pure $Na_2CO_3$ in 1.0 dm3 solution = Molarity $\times$ Molar mass $\times$ Volume (in dm3) Mass of pure $Na_2CO_3$ = $0.045 M \times 106 g/mol \times 1.0 dm^3 = 4.77 g$ Percentage purity = $\frac{\text{Mass of pure } Na_2CO_3}{\text{Mass of impure sample}} \times 100\%$ Percentage purity = $\frac{4.77 g}{10.0 g} \times 100\% = 47.7\%$ Water of Crystallization: This involves determining 'x' in a hydrated salt formula (e.g., $Na_2CO_3.xH_2O$). Worked Example 3 (Water of Crystallization): 2.86 g of $Na_2CO_3.xH_2O$ was dissolved to make 250 cm3 of solution. 25.0 cm3 of this solution required 20.0 cm3 of 0.1 M HCl for complete neutralization. Redox titrations involve the transfer of electrons between the reactants. One reactant is oxidized (loses electrons), and the other is reduced (gains electrons). Common oxidizing agents include potassium permanganate ($KMnO_4$) and potassium dichromate ($K_2Cr_2O_7$). Potassium permanganate acts as a self-indicator because its purple colour disappears as it is reduced to colourless $Mn^{2+}$ ions. The end point is reached when a permanent faint pink colour appears. This type of titration is commonly used to determine the concentration of iron(II) salts or oxalates. Example Redox Reaction (Permanganate and Iron(II)): $MnO_4^-(aq) + 8H^+(aq) + 5Fe^{2+}(aq) \rightarrow Mn^{2+}(aq) + 5Fe^{3+}(aq) + 4H_2O(l)$ Here, 1 mole of $MnO_4^-$ reacts with 5 moles of $Fe^{2+}$. The same titration formula $\frac{C_A V_A}{C_B V_B} = \frac{n_A}{n_B}$ can be applied, with $n_A$ and $n_B$ being the stoichiometric coefficients from the balanced redox equation. Quantitative analysis involves determining the amount or concentration of a substance in a sample. For SS3, the focus is primarily on Volumetric Analysis (Titration).
Water Quality Testing (Nigerian context): Application: Quantitative and qualitative analysis is extensively used in Nigeria to assess the potability of drinking water, especially from boreholes, wells, and sachet water production plants.
Relevance: Titration can determine the concentration of hardness-causing ions (Ca2+, Mg2+), chloride levels, or alkalinity. Qualitative tests can detect heavy metal contamination (e.g., lead ions, which might dissolve from old pipes) or the presence of specific anions like nitrates (from agricultural run-off) that can pose health risks. This ensures water meets standards set by NAFDAC and protects public health. Food Authentication and Adulteration Detection: Application: Analytical techniques are vital in the Nigerian food industry to ensure product quality and detect common adulterations.
Relevance: Qualitative tests for starch can identify if garri or semovita has been contaminated with foreign starches. Tests for fats/oils are used to check the purity of palm oil or groundnut oil, detecting if they have been mixed with cheaper, inferior oils. Protein tests are used in milk and other protein-rich foods to ensure nutritional content and absence of diluents. Quantitative analysis can determine the exact fat content in a snack or protein in a local food product.
Soil Analysis in Agriculture: Application: Farmers in Nigeria can benefit from soil analysis to optimize crop yields.
Relevance: Titration is used to determine soil pH (acidic, neutral, alkaline), which influences nutrient availability for crops like yam, cassava, or maize. Qualitative tests can identify the presence of essential mineral ions or harmful contaminants, helping farmers select appropriate fertilizers and amendments, thereby boosting food security.