Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Question A basic buffer contains: NH 3 = 0.10  mol , NH 4 + = 0.10  mol Strong acid added: HCl = 0.05  mol Given: p K b ( NH 3 ) = 4.745 Find the change in pH after adding HCl. ✍️ Short Solution This is a weak base buffer (NH₃–NH₄Cl). When HCl is added, it reacts completely with NH₃, converting it into NH₄⁺. We then apply the Henderson–Hasselbalch equation for basic buffers to find the new pH. 🔹 Step 1 — Buffer System NH 3 + NH 4 + \text{NH}_3 + \text{NH}_4^+ Both initially = 0.10 mol → Perfect weak base buffer. 🔹 Step 2 — Henderson–Hasselbalch Equation (Basic Buffer) pOH = p K b + log ⁡ ( salt base ) \text{pOH} = pK_b + \log\left(\frac{\text{salt}}{\text{base}}\right) pH = 14 − pOH \text{pH} = 14 - \text{pOH} Given: p K b = 4.745 🔹 Step 3 — Reaction With Strong Acid HCl reacts fully with NH₃: NH 3 + HCl → NH 4 + \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4^+ HCl added = 0.05 mol New moles: NH 3 = 0.10 − 0.05 = 0.05  mol \text...

  ❓ Question FOR: If for θ ∈ [ − π 3 ,   0 ] , ( x , y ) = ( 3 tan ⁡  ⁣ ( θ + π 3 ) ,   2 tan ⁡  ⁣ ( θ + π 6 ) ) lie on the curve x y + α x + β y + γ = 0 , xy + \alpha x + \beta y + \gamma = 0, then the value of α 2 + β 2 + γ 2 \alpha^2 + \beta^2 + \gamma^2 is equal to ? 🖼️ Question Image ✍️ Short Solution We are told that for every θ \theta  in the interval, x = 3 tan ⁡ ( θ + π 3 ) , y = 2 tan ⁡ ( θ + π 6 ) x = 3\tan\left(\theta + \frac{\pi}{3}\right), \quad y = 2\tan\left(\theta + \frac{\pi}{6}\right) always satisfies x y + α x + β y + γ = 0. xy + \alpha x + \beta y + \gamma = 0. That means there is a fixed relation between x x x and y y y which does not involve θ \theta . We’ll eliminate θ \theta  using a standard tangent identity. 🔹 Step 1 — Define new angles and variables Let A = θ + π 3 , B = θ + π 6 . A = \theta + \frac{\pi}{3}, \quad B = \theta + \frac{\pi}{6}. Then, x = 3 tan ⁡ A , y = 2 tan ⁡ B . x = 3\tan A,\quad y = 2\tan B. Compu...

  ❓ Question Dumas Method – Nitrogen % in 60 Sec! A quick, exam-oriented breakdown of how Dumas Method helps you find percentage of nitrogen in an organic compound using only the nitrogen gas collected . 📝 1️⃣ Dumas Method Concept Organic compound is heated strongly → Nitrogen present in compound converts into N₂ gas → This N₂ is collected over water and its volume, temperature, pressure are measured. ✔ No chemical titration ✔ 100% physical calculation ✔ Based on ideal gas equation 📝 2️⃣ Correcting the Pressure (MOST Important Step) Gas is collected over water , so total pressure = Pressure of dry N₂ + water vapour pressure. Corrected pressure: P N 2 = P total − P aq. tension​ Given: P total = 715  mmHg , P aq. tension = 15  mmHg So, P N 2 = 700  mmHg 📝 3️⃣ Ideal Gas Equation (Find moles of N₂) P V = n R T Convert units: P = 700  mmHg = 700 / 760  atm P = 700\ \text{mmHg} = 700/760\ \text{atm} V = 50  mL = 0.0...

  ❓ Question Let the set of all values of p ∈ R p \in \mathbb{R}  for which both the roots of the equation x 2 − ( p + 2 ) x + ( 2 p + 9 ) = 0 are negative real numbers be the interval ( α , β ] (\alpha, \beta] . Then the value of β − 2 α is equal to ? 🖼️ Question Image ✍️ Short Solution We have a quadratic: x 2 − ( p + 2 ) x + ( 2 p + 9 ) = 0 Compare with x 2 − S x + P = 0 x^2 - Sx + P = 0 , where Sum of roots = S = p + 2 = S = p+2 Product of roots = P = 2 p + 9 = P = 2p+9 For both roots to be negative real numbers , we need: Real roots → Discriminant Δ ≥ 0 Both roots negative → Sum of roots < 0 <0 Product of roots > 0 >0 🔹 Step 1 — Conditions from sum and product Sum < 0: p + 2 < 0 ⇒ p < − 2 Product > 0: 2 p + 9 > 0 ⇒ p > − 9 2​ Together: − 9 2 < p < − 2 🔹 Step 2 — Discriminant condition Δ = ( p + 2 ) 2 − 4 ( 2 p + 9 ) ≥ 0 Compute: Δ = ( p 2 + 4 p + 4 ) − ( 8 p + 36 ) = p 2 − 4 p − 32 So...

  ❓ Concept Limiting Reagent & Water Volume Trick in 60 Seconds! How do you quickly find how much water forms in a combustion reaction without writing long stoichiometry tables ? This 60-second hack covers: ✔ Limiting reagent shortcut ✔ Quick mole-ratio conversion ✔ Mass → volume trick for liquid water ✍️ Short Explanation Let’s break the entire idea into three ultra-fast steps , using the popular combustion example: C 4 H 10 + 13 2 O 2 → 4 C O 2 + 5 H 2 O \mathrm{C_4H_{10}} + \frac{13}{2} \mathrm{O_2} \rightarrow 4\mathrm{CO_2} + 5\mathrm{H_2O} This trick works for any reaction that forms liquid water . 🔹 Step 1 — Identify the Limiting Reagent FAST Instead of computing everything, compare “ moles available : moles required ”. Example: 174 kg butane → 174000 58 ≈ 3000 \frac{174000}{58} \approx 3000 320 kg O₂ → 320000 32 = 10000 \frac{320000}{32} = 10000 Reaction needs 6.5 mol O₂ per mol C₄H₁₀ . Demand for O₂ to consume all butane: 3000 × 6.5 = 19500 ...

  ❓ Question Evaluate the integral: ∫ 0 π ( x + 3 ) sin ⁡ x 1 + 3 cos ⁡ 2 x   d x 🖼️ Question Image ✍️ Short Solution We split the integral into two parts: I = ∫ 0 π x sin ⁡ x 1 + 3 cos ⁡ 2 x   d x + 3 ∫ 0 π sin ⁡ x 1 + 3 cos ⁡ 2 x   d x . I = \int_{0}^{\pi} \frac{x\sin x}{1+3\cos^{2}x}\,dx + 3\int_{0}^{\pi} \frac{\sin x}{1+3\cos^{2}x}\,dx. Let: I 1 = ∫ 0 π x sin ⁡ x 1 + 3 cos ⁡ 2 x   d x , I 2 = ∫ 0 π sin ⁡ x 1 + 3 cos ⁡ 2 x   d x . I_1=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx,\qquad I_2=\int_0^\pi \frac{\sin x}{1+3\cos^2x}\, dx. 🔹 Step 1 — Simplify I 1 I_1 ​ using substitution symmetry Let J = ∫ 0 π x sin ⁡ x 1 + 3 cos ⁡ 2 x   d x . J=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx. Substitute: x = π − t , d x = − d t , sin ⁡ ( π − t ) = sin ⁡ t , cos ⁡ ( π − t ) = − cos ⁡ t . x=\pi - t,\qquad dx=-dt,\qquad \sin(\pi - t)=\sin t,\qquad \cos(\pi - t)=-\cos t. Thus: J = ∫ 0 π ( π − t ) sin ⁡ t 1 + 3 cos ⁡ 2 t   d t . J=\int_0^\pi \frac{(\pi - t)\sin t}{1+3\cos^2 t}\,...

  ❓ Question Why is Al³⁺ stable , Tl⁺ highly stable , and Tl³⁺ a strong oxidising agent? Let’s break the entire concept in one crisp explanation. ✍️ Short Explanation The inert pair effect refers to the reluctance of the ns² electrons (the "inert pair") to participate in bonding for heavy p-block elements . This means as we go down a group, especially in groups 13–16 , the lower oxidation state becomes more stable , and the higher oxidation state becomes less stable . Let's apply this to Group 13: Al → Ga → In → Tl . 🔹 1. Why Al³⁺ is Stable Aluminium is in the 3rd period . Relativistic effects are very small. The 3s² electrons are not inert — they easily participate in bonding. So Al prefers oxidation state +3 (Al³⁺). 👉 Al³⁺ is very stable, does not get reduced easily. 🔹 2. Why Tl⁺ is More Stable than Tl³⁺ Thallium (Tl) is in the 6th period . The 6s² electrons are held tightly due to relativistic contraction. They behave like ...

  ❓ Question The number of paramagnetic metal complex species among the following — that have the same number of unpaired electrons — is asked for: [ Co(NH 3 ) 6 ] 3 + ,    [ Co(C 2 O 4 ) 3 ] 3 − ,    [ MnCl 6 ] 3 − ,    [ Mn(CN) 6 ] 3 − ,    [ CoF 6 ] 3 − ,    [ Fe(CN) 6 ] 3 − ,    [ FeF 6 ] 3 − We must find which of these are paramagnetic and then see how many of them share the same number of unpaired electrons. 🖼️ Question Image ✍️ Short Solution (stepwise) We analyse each complex: General steps: Determine metal oxidation state from complex charge. Write the d-electron count. Decide whether ligand is strong-field (low-spin) or weak-field (high-spin). Count unpaired electrons in octahedral splitting. 1. [ Co(NH 3 ) 6 ] 3 + [ \text{Co(NH}_3)_6 ]^{3+} Co oxidation: + 3 +3  → d 6 d^6 . NH₃ is a borderline/relatively strong ligand for Co³⁺ → typically low-spin . Low-spin d 6 d^6 d 6 → t 2 g 6 e g 0 t_{2g}^6 e_g^0  → 0 unpaired → di...