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Chemistry · Ionic Equilibrium

JEE Advanced 2018 — Paper 1 — Question 25

The solubility of a salt of weak acid ( AB\mathbf{A B} ) at pH 3 is Y×10−3 mol L−1\mathbf{Y} \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}. The value of Y\mathbf{Y} is ____\_\_\_\_ .

(Given that the value of solubility product of AB(Ksp)=2×10−10\mathbf{A B}\left(\mathrm{K}_{\mathrm{sp}}\right)=2 \times 10^{-10} and the value

of ionization constant of HB (Ka)=1×10−8)\left.\left(K_{a}\right)=1 \times 10^{-8}\right)

Answer: 4.47

Numerical answer — enter this value.

Step-by-step solution

AB⇌As++B−−(S−x)\mathrm{AB} \rightleftharpoons \mathrm{A}_{\mathrm{s}}^{+}+ \mathrm{B}^- \underset{(\mathrm{S}-\mathrm{x})} {-}

2×10−10=S(S−x)…(1)2 \times 10^{-10}=S(S-x) …(1)

B−(S−x)+H+10−3⇌xHBC\underset{(\mathrm{S}-\mathrm{x})}{\mathrm{B}^{-}}+\underset{10^{-3}}{\mathrm{H}^{+}} \rightleftharpoons \underset{\mathrm{x}}{ } \mathrm{HB}^{\mathrm{C}}

110−8=x(S−x)×10−3\frac{1}{10^{-8}}=\frac{\mathrm{x}}{(\mathrm{S}-\mathrm{x}) \times 10^{-3}}

xS−x=105…(2)\frac{x}{S-x}=10^{5} …(2)

Multiply equation (1) and (2).

S. x=2×10−5x=2 \times 10^{-5}

From Eq. (1)

S2−Sx=2×10−10\mathrm{S}^{2}-\mathrm{Sx}=2 \times 10^{-10}

S2−2×10−5=2×10−10S^{2}-2 \times 10^{-5}=2 \times 10^{-10}

S2=2×10−5+2×10−10≅2×10−5\mathrm{S}^{2}=2 \times 10^{-5}+2 \times 10^{-10} \cong 2 \times 10^{-5}

S=4.47×10−3\mathrm{S}=4.47 \times 10^{-3}

y=4.47y=4.47

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 1
Subject
Chemistry
Chapter
Ionic Equilibrium
Topic
Sparingly Soluble Salts, Solubility Product & Precipitation Conditions
The solubility of a salt of weak acid ( A B ) at pH 3 is Y × 10 -3… | JEE Advanced 2018 PYQ with Solution · DhiX AI