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Chemistry · Ionic Equilibrium

JEE Main 2025 — 23 January, Morning Shift — Question 45

If 1 mM solution of ethylamine produces pH=9\mathrm{pH}=9, then the ionization constant (Kb)\left(\mathrm{K}_{\mathrm{b}}\right) of ethylamine is 10−x10^{-x}. The value of x is___ (nearest integer).[0pt] [The degree of ionization of ethylamine can be neglected with respect to unity.]

Answer: 7

Numerical answer — enter this value.

Step-by-step solution

C2H5NH2(aq)+H2O⇌C2H2NH3++O⊖H\quad \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{2} \mathrm{NH}_{3}^{+}+\stackrel{\ominus}{\mathrm{O}} \mathrm{H}

C=10−3M\mathrm{C}=10^{-3} \mathrm{M}

C(1−α)\mathrm{C}(1-\alpha)

⇒C=10−3=10−5=10−5\Rightarrow \mathrm{C}=10^{-3} \quad=10^{-5} \quad=10^{-5}

1−α≃11-\alpha \simeq 1

Given, PH=9⇒POH=5⇒[O⊖H]=10−5M\mathrm{P}^{\mathrm{H}}=9 \Rightarrow \mathrm{P}^{\mathrm{OH}}=5 \Rightarrow[\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]=10^{-5} \mathrm{M}

Now, Kb=[C2H5NH3+][O⊖H][C2H5NH2]\mathrm{K}_{\mathrm{b}}=\frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right][\stackrel{\ominus}{\mathrm{O}} \mathrm{H}]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}

⇒Kb=10−5×10−510−3=10−7\Rightarrow \mathrm{K}_{\mathrm{b}}=\frac{10^{-5} \times 10^{-5}}{10^{-3}}=10^{-7}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
23 January, Morning Shift
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
pH of solution containing implicit reaction

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