Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Ionic Equilibrium question

2025 · 23 Jan · Shift 1 · Q23
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Ionic Equilibrium
  5. /2025 · 23 Jan · Shift 1 · Q23

Ionic Equilibrium question

2025 · 23 Jan · Shift 1 · Q23

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
If 1 mM solution of ethylamine produces pH=9\mathrm{pH}=9pH=9, then the ionization constant (Kb)\left(\mathrm{K}_{\mathrm{b}}\right)(Kb​) of ethylamine is 10−x10^{-x}10−x. The value of xxx is ‾\underline{\hspace{2cm}}​ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]
Numerical answer
View written solutionFree

Correct answer: 7

  1. Given data

    • Concentration of ethylamine: C=1 mM=10−3 MC = 1\,\text{mM} = 10^{-3}\,\text{M}C=1mM=10−3M
    • pH of solution: pH=9\text{pH} = 9pH=9

  2. Find pOH and hydroxide ion concentration pOH=14−9=5\text{pOH} = 14 - 9 = 5pOH=14−9=5 Therefore, [OH−]=10−5 M[\text{OH}^-] = 10^{-5}\,\text{M}[OH−]=10−5M

  3. Write the base ionization equilibrium Ethylamine is a weak base: C2H5NH2+H2O⇌C2H5NH3++OH−\mathrm{C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-}C2​H5​NH2​+H2​O⇌C2​H5​NH3+​+OH−

    Let the initial concentration be C=10−3C = 10^{-3}C=10−3 M and degree of ionization be small. Then at equilibrium:

    • [OH−]=x=10−5[\mathrm{OH^-}] = x = 10^{-5}[OH−]=x=10−5
    • [C2H5NH3+]=x=10−5[\mathrm{C_2H_5NH_3^+}] = x = 10^{-5}[C2​H5​NH3+​]=x=10−5
    • [C2H5NH2]≈C=10−3[\mathrm{C_2H_5NH_2}] \approx C = 10^{-3}[C2​H5​NH2​]≈C=10−3

  4. Expression for KbK_bKb​ Kb=[C2H5NH3+][OH−][C2H5NH2]K_b = \frac{[\mathrm{C_2H_5NH_3^+}][\mathrm{OH^-}]}{[\mathrm{C_2H_5NH_2}]}Kb​=[C2​H5​NH2​][C2​H5​NH3+​][OH−]​

    Substituting values: Kb=(10−5)(10−5)10−3K_b = \frac{(10^{-5})(10^{-5})}{10^{-3}}Kb​=10−3(10−5)(10−5)​ Kb=10−7K_b = 10^{-7}Kb​=10−7

  5. Compare with given form Kb=10−xK_b = 10^{-x}Kb​=10−x So, x=7x = 7x=7

  6. Nearest integer 7\boxed{7}7​

PreviousNext

More from Ionic Equilibrium

  • pH of water is 7 at 25∘C. If water is heated to 80∘C., it's pH will :2025 · MCQ
  • Ksp​ for Cr(OH)3​ is 1.6×10−30. What is the molar solubility of this salt in water?2025 · MCQ
  • The observed and normal molar masses of compound MX2​ are 65.6 and 164 respectively. The percent degree of ionisation of MX2​ is ​%. (Nearest integer)2025 · Numerical
  • A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH - pKa​)?2025 · MCQ
  • Arrange the following in increasing order of solubility product : Ca(OH)2​,AgBr,PbS,HgS2025 · MCQ
  • Ka​ for CH3​COOH is 1.8×10−5 and Kb​ for NH4​OH is 1.8×10−5. The pH of ammonium acetate solution will be ​…2024 · Numerical
  • Solubility of calcium phosphate (molecular mass, M) in water is Wg​ per 100 mL at 25∘C. Its solubility product at 25∘C will be approximately.2024 · MCQ
  • Consider the dissociation of the weak acid HX as given below HX(aq)⇌H+(aq)+X−(aq),Ka=1.2×10−5[Ka​: dissociation…2024 · Numerical