One litre buffer solution was prepared by adding 0.10text mol each of textNH_3 and textNH_4textCl in deionised water. The change in pH on addition of 0.05text mol of textHCl to the above solution is dots times 10^-2 (Nearest integer) [cite: 433, 434] Given: textpK_textb of textNH_3 = 4.745 and log_103 = 0.477

Numerical Answer Type:
Enter a numerical value Answer: 47.5 to 48.5 +4 marks

Solution & Explanation

### Related Formula textpOH = textpK_textb + log frac[textSalt][textBase] textpH = 14 - textpOH ### Core Logic Initially, the basic buffer solution contains: [textSalt] = [textNH4^+] = 0.10text mol, quad [textBase] = [textNH3] = 0.10text mol textpOHtextinitial = 4.745 + log frac0.100.10 = 4.745 When 0.05text mol of strong acid textHCl is introduced, it reacts stoichiometrically with the weak base textNH_3: [cite: 1049, 1050] beginarrayrcccc & textNH3 & + & textH^+ & ightarrow & textNH4^+ \ textInitial (mol): & 0.10 & & 0.05 & & 0.10 \ textFinal (mol): & 0.05 & & 0 & & 0.15 endarray ### Step 1: Computing Post-Acid pOH and pH Recalculating via Henderson's equation: textpOHtextfinal = 4.745 + log frac0.150.05 = 4.745 + log 3 The total shift value follows as: Delta textpOH = textpOHtextfinal - textpOHtextinitial = log 3 = 0.477 Since textpH = 14 - textpOH: Delta textpH = -Delta textpOH = -0.477 Expressing the structural magnitude in scientific notation format: |Delta textpH| = 0.477 = 47.7 times 10^-2 approx 48 times 10^-2 ### Pattern Recognition Buffer shifting rule: Adding an acid consumes base and builds salt. The base drops from 0.1 to 0.05 (halved), while salt grows from 0.1 to 0.15 (tripled). The ratio flips to 3, introducing a clean log 3 change factor into the solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

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Ionic Equilibrium

More Ionic Equilibrium Previous-Year Questions — Page 6

Q62 jee_main_2024_31_jan_morning Equilibrium Constant
For the given reaction, choose the correct expression of K_C from the following :- Fe_(aq)^3+ + SCN_(aq)^- rightleftharpoons (FeSCN)_(aq)^2+
  • A. K_C = frac[FeSCN^2+][Fe^3+][SCN^-]
  • B. K_C = frac[Fe^3+][SCN^-][FeSCN^2+]
  • C. K_C = frac[FeSCN^2+][Fe^3+]^2[SCN^-]^2
  • D. K_C = frac[FeSCN^2+]^2[Fe^3+][SCN^-]

Solution

### Related Formula K_C = frac[textProducts][textReactants] ### Core Logic K_C = fractextProducts ion conc.textReactants ion conc. K_C = frac[FeSCN^2+][Fe^3+][SCN^-] ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

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