An aqueous solution of mathrmHCl with mathrmpH\ 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The mathrmpH of mathrmHCl solution would: (Given log 2 = 0.30)

Solution & Explanation

### Related Formula textpH = -log_10[textH^+] ### Core Logic For the initial solution: textpH = 1.0 implies [textH^+]_1 = 10^-1 = 0.1 text M When we dilute the solution by adding an equal volume of water, the final volume is doubled (V_2 = 2V_1). Thus, the final concentration is halved: [textH^+]_2 = frac[textH^+]_12 = frac0.12 = 0.05 text M Now, calculate the new textpH: textpH_2 = -log_10(0.05) = -log_10left(frac120right) = log_10 20 = log_10(10 times 2) = 1 + log_10 2 textpH_2 = 1 + 0.30 = 1.30 ### Pattern Recognition Diluting any strong acid by 2 times increases the textpH by exactly log_10 2 approx 0.30. Thus, 1.0 + 0.3 = 1.3 immediately. ### 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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