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Given below are two statements : Statement (I): Potassium hydrogen phthalate is a primary standard for standardisation of sodium hydroxide solution. Statement (II) : In this titration phenolphthalein can be used as indicator. In the light of the above statements, choose the most appropriate answer from the options given below:

Solution & Explanation

### Core Logic Statement (I): Potassium hydrogen phthalate (KHP) is widely used as a primary standard in analytical chemistry for standardizing strong bases like NaOH. This is because it is highly pure, non-hygroscopic, stable, and has a relatively high molar mass, making its concentration reliable and stable over time. Statement (II): KHP is a weak acid and NaOH is a strong base. The titration of a weak acid with a strong base yields an equivalence point in the weakly basic range (pH > 7). Phenolphthalein changes colour in the pH range 8.3 to 10.0, making it the perfect indicator for this titration. ### Step 1: Evaluate Statements Statement I is correct. Statement II is correct. ### Pattern Recognition Weak Acid vs Strong Base rightarrow Equivalence pH > 7 rightarrow Phenolphthalein is the indicator of choice. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium Class 11 Chemistry: Some Basic Concepts of Chemistry

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Some Basic Concepts Of Chemistry

More Some Basic Concepts of Chemistry Previous-Year Questions — Page 4

Q89 jee_main_2024_01_february_morning Stoichiometry
Consider the following reaction: 3PbCl_2 + 2(NH_4)_3PO_4 rightarrow Pb_3(PO_4)_2 + 6NH_4Cl If 72 mathrm~mmol of PbCl_2 is mixed with 50 mathrm~mmol of (NH_4)_3PO_4, then amount of Pb_3(PO_4)_2 formed is ... mmol. (nearest integer)
Numerical Answer. Answer: 24 to 24

Solution

### Related Formula textMoles of Product = textMoles of Limiting Reagent times fractextStoichiometry of ProducttextStoichiometry of Limiting Reagent ### Core Logic From the balanced chemical equation: 3 text moles of PbCl_2 text react with 2 text moles of (NH_4)_3PO_4. Let's find the limiting reagent (L.R.) by dividing given millimoles by stoichiometric coefficients: For PbCl_2: frac723 = 24 For (NH_4)_3PO_4: frac502 = 25 Since 24 < 25, PbCl_2 is the limiting reagent and will completely consume. ### Step 1: Calculate Product Moles Moles of Pb_3(PO_4)_2 formed depends entirely on PbCl_2. 3 mmol of PbCl_2 produces 1 mmol of Pb_3(PO_4)_2. Therefore, 72 mmol of PbCl_2 will produce: frac13 times 72 = 24 mathrm~mmol of Pb_3(PO_4)_2. ### Pattern Recognition Always identify the Limiting Reagent by taking the ratio n / textcoefficient. The smallest ratio dictates the extent of the reaction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q85 jee_main_2024_29_january_evening Volumetric Titration and Molarity
If 50text mL of 0.5text M oxalic acid is required to neutralise 25text mL of mathrmNaOH solution, the amount of mathrmNaOH in 50text mL of given mathrmNaOH solution is ________ g.
Numerical Answer. Answer: 4 to 4

Solution

### Related Formula textEquivalents of Acid = textEquivalents of Base N_1 V_1 = N_2 V_2 implies (M_1 times n_1) times V_1 = (M_2 times n_2) times V_2 ### Core Logic For oxalic acid (textH_2textC_2textO_4), the valence factor (n-factor) is 2. For textNaOH, the n-factor is 1. Substituting the values into the normality equivalence expression: 50 times 0.5 times 2 = 25 times M_textNaOH times 1 50 = 25 times M_textNaOH implies M_textNaOH = 2text M ### Step 1: Mass Isolation To find the mass of textNaOH present in 50text mL of this solution: textMass = textMolarity times textVolume (in L) times textMolar Mass textMass = 2 times left(frac50, 1000right) times 40 = 2 times 0.05 times 40 = 4text g ### Pattern Recognition Remember to use the correct n-factor (2) for dibasic oxalic acid during equivalence matching to avoid calculation errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q83 jee_main_2024_27_jan_morning Stoichiometry
Mass of methane required to produce 22text g of textCO_2 after complete combustion is textquadquad g. (Given Molar mass in textg mol^-1: textC=12.0, textH=1.0, textO=16.0)
Numerical Answer. Answer: 8 to 8

Solution

### Related Formula Balanced combustion chemical equation: textCH_4 + 2textO_2 rightarrow textCO_2 + 2textH_2textO textMoles = fractextMasstextMolar Mass ### Step 1: Determine moles of product generated textMolar Mass of CO_2 = 12 + (2 times 16) = 44text g mol^-1 textMoles of CO_2 text produced = frac2244 = 0.5text moles ### Step 2: Relate to input mass via stoichiometry metrics From the balanced equation, 1text mole of CH_4 produces 1text mole of CO_2. textRequired Moles of CH_4 = 0.5text moles textMolar Mass of CH_4 = 12 + (4 times 1) = 16text g mol^-1 textMass of CH_4 = 0.5 times 16 = 8text g ### Pattern Recognition 22text g of CO_2 is exactly half a mole. By stoichiometry ratios, half a mole of methane is needed, which translates to 8text g. ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry
Q88 jee_main_2024_30_jan_morning Mole Concept
0.05text cm thick coating of silver is deposited on a plate of 0.05text m^2 area. The number of silver atoms deposited on plate are ________ times 10^23. (At mass Ag=108, d=7.9text g cm^-3)
Numerical Answer. Answer: 11 to 11

Solution

### Related Formula textVolume = textArea times textThickness textMass = textDensity times textVolume textMoles = fractextMasstextMolar Mass textNumber of Atoms = textMoles times N_A ### Step 1: Calculate Volume of Coating Area = 0.05 text m^2 = 0.05 times 10^4 text cm^2 = 500 text cm^2 Thickness = 0.05 text cm Volume = 500 text cm^2 times 0.05 text cm = 25 text cm^3 ### Step 2: Calculate Mass and Moles Mass = Volume times Density = 25 text cm^3 times 7.9 text g/cm^3 = 197.5 text g Moles of Ag = frac197.5108 = 1.8287 text moles ### Step 3: Calculate Number of Atoms textNumber of Atoms = 1.8287 times 6.022 times 10^23 = 11.01 times 10^23 Rounding to nearest integer, we get 11. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry Class 12 Chemistry: Electrochemistry

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