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Ionic Equilibrium question

2018 · Shift 1 · Q9
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Ionic Equilibrium question

2018 · Shift 1 · Q9

JEE AdvancedChemistryIonic EquilibriumNumerical+3 / −1
The solubility of a salt of weak acid (AB)(AB)(AB) at pH 3pH\,3pH3 is Y×10−3mol L−1.Y \times {10^{ - 3}}mol\,{L^{ - 1}}.Y×10−3molL−1. The value of YYY is ‾\underline{\hspace{2cm}}​. (Given that the value of solubility product of ABABAB (Ksp)=2×10−10\left( {{K_{sp}}} \right) = 2 \times {10^{ - 10}}(Ksp​)=2×10−10 and the value of ionization constant of HBHBHB(Ka)=1×10−8\left( {{K_a}} \right) = 1 \times {10^{ - 8}}(Ka​)=1×10−8)
Numerical answer
View written solutionFree

Correct answer: 4.47

  1. Dissolution equilibrium of the salt

For the salt ABABAB: AB(s)⇌A++B−AB(s) \rightleftharpoons A^+ + B^-AB(s)⇌A++B−

Given: Ksp=[A+][B−]=2×10−10K_{sp} = [A^+][B^-] = 2\times 10^{-10}Ksp​=[A+][B−]=2×10−10

  1. Acid-base equilibrium of the anion B−B^-B−

Since HBHBHB is a weak acid: HB⇌H++B−HB \rightleftharpoons H^+ + B^-HB⇌H++B− Ka=[H+][B−][HB]=10−8K_a = \frac{[H^+][B^-]}{[HB]} = 10^{-8}Ka​=[HB][H+][B−]​=10−8

At pH=3pH=3pH=3, [H+]=10−3[H^+] = 10^{-3}[H+]=10−3

Hence, [HB][B−]=[H+]Ka=10−310−8=105\frac{[HB]}{[B^-]} = \frac{[H^+]}{K_a} = \frac{10^{-3}}{10^{-8}} = 10^5[B−][HB]​=Ka​[H+]​=10−810−3​=105

So most of the dissolved BBB exists as HBHBHB.

  1. Relate total solubility to free B−B^-B−

Let solubility be sss mol L−1^{-1}−1. Then: [A+]=s[A^+] = s[A+]=s

Total concentration of species derived from BBB is also sss: s=[B−]+[HB]s = [B^-] + [HB]s=[B−]+[HB]

Using [HB]=[H+]Ka[B−]=105[B−][HB] = \frac{[H^+]}{K_a}[B^-] = 10^5[B^-][HB]=Ka​[H+]​[B−]=105[B−]

Thus, s=[B−](1+105)s = [B^-](1+10^5)s=[B−](1+105)

So, [B−]=s1+105[B^-] = \frac{s}{1+10^5}[B−]=1+105s​

  1. Apply solubility product

Ksp=[A+][B−]=s⋅s1+105K_{sp} = [A^+][B^-] = s\cdot \frac{s}{1+10^5}Ksp​=[A+][B−]=s⋅1+105s​

Therefore, s21+105=2×10−10\frac{s^2}{1+10^5} = 2\times 10^{-10}1+105s2​=2×10−10

Since 1+105=100001≈1051+10^5 = 100001 \approx 10^51+105=100001≈105, s2≈2×10−10×105=2×10−5s^2 \approx 2\times 10^{-10}\times 10^5 = 2\times 10^{-5}s2≈2×10−10×105=2×10−5

So, s=2×10−5=2×10−2.5s = \sqrt{2\times 10^{-5}} = \sqrt{2}\times 10^{-2.5}s=2×10−5​=2​×10−2.5

Now, 10−2.5=10−31010^{-2.5} = 10^{-3}\sqrt{10}10−2.5=10−310​

Hence, s=210×10−3=20×10−3s = \sqrt{2}\sqrt{10}\times 10^{-3} = \sqrt{20}\times 10^{-3}s=2​10​×10−3=20​×10−3

s≈4.47×10−3 mol L−1s \approx 4.47\times 10^{-3}\,\text{mol L}^{-1}s≈4.47×10−3mol L−1

  1. Find YYY

Given solubility is Y×10−3Y\times 10^{-3}Y×10−3 mol L−1^{-1}−1, Y=4.47Y = 4.47Y=4.47

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