JEE MainChemistryModeratePhysical Chemistry

On mixing equal volumes of 0.1 M AgNO3 & 0.8 M KCN solutions,
what of original Ag+ is in the form of free Ag+ ions ? Given: Kd of
[Ag(CN)2]- = 10-18. Report your answer after multiplying it by 4.5 ×
1016. (Assume no hydrolysis of any ion)

Show answer and solution

10^{- 18} = \frac{\left\lbrack {Ag}^{+} \right\rbrack_{free}(0.3)^{2}}{(0.05)} \Rightarrow \left\lbrack {Ag}^{+} \right\rbrack_{free} = \frac{5}{9} \times 10^{- 18}M

\%\ of\ {Ag}^{+}\left( {Ag}^{+}\% \right) = \frac{\frac{5}{9} \times 10^{- 18}}{0.05} \times 100 \times 4.5 \times 10^{16} = 50\%

Source: Grass-G / CH-23 Chemistry Paper 31 Dec. FST.docx

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