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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Chemistry
Chapter
Ionic Equilibrium

BITSAT 2014 · Chemistry · Ionic Equilibrium

K_ c for the the reaction, [Ag(CN)_ 2 ] leftharpoons Ag^ + +2 CN^ -...

K_{c}forthethereaction,[Ag(CN)2]⇌Ag++2CN−,for the the reaction, \left[\mathrm{Ag}(\mathrm{CN})_{2}\right] \rightleftharpoons \mathrm{Ag}^{+}+2 \mathrm{CN}^{-}, the equillibrium constant at 25∘C25^{{}^{\circ}} \mathrm{C} is 4.0×10−194.0 \times 10^{-19} then the silver ion concentration in a solution which was originally 0.1 molar in KCN\mathrm{KCN} and 0.03 molar in AgNO3\mathrm{AgNO}_{3} is:

Options

  1. A7.5×10187.5 \times 10^{18}
  2. B7.5×10−197.5 \times 10^{-19}
  3. C7.5×10197.5 \times 10^{19}
  4. D7.5×10−187.5 \times 10^{-18}

Correct answer

D. 7.5×10−187.5 \times 10^{-18}

Step-by-step solution

\left.\begin{array}{c}2 \mathrm{KCN}+\mathrm{AgNO}=\left[\underset{0.1}{0.03}=\left[\begin{array}{c}\mathrm{Ag}(\mathrm{CN})_{2} \\ 0\end{array}\right]^{-}\right. \\ (0.1-0.06) & 0\end{array}\right] KNO3+K0+(0.03)\begin{array}{c}\mathrm{KNO}_{3}+\mathrm{K}_{0}^{+} \\ (0.03)\end{array} ∴[Ag(CN)2]−=Ag++0.03aa\therefore\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}=\mathrm{Ag}^{+}+\underset{\mathrm{a}}{0.03} \mathrm{a} \quad2 CN(\mathrm{CN}(0.03-a)a.\mathrm{a}) \quad \mathrm{a} .04(leftfromKCN\mathrm{left} \mathrm{from} \mathrm{KCN}\quad0.04 ≈\approx0.04\therefore \quad Kc=\mathrm{K}_{\mathrm{c}}=4 ×\times10^{-19}=(0.04)2×a0.03a=\frac{(0.04)^{2} \times \mathrm{a}}{0.03}\mathrm{a}=7.5 ×\times10^{-18}
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