Ionic EquilibriumHard
Question
An amount of 0.01 moles of solid AgCN is rendered soluble in 1 L by adding just sufficient excess cyanide ion to form Ag(CN)2− and the concentration of free cyanide ion is 2.5 × 10–7 M. Determine [Ag+] in the solution neglecting hydrolysis of cyanide ion. The value of Kdiss for Ag(CN)2− = 1.0 × 10–20.
Options
A.6.25 × 10−9 M
B.1.6 × 10−9 M
C.1.6 × 10−7 M
D.6.25 × 10−7 M
Solution
$\left\lbrack Ag(CN)_{2}^{-} \right\rbrack = 0.01\text{ M}$
$K_{diss} = \frac{\left\lbrack Ag^{+} \right\rbrack\left\lbrack CN^{-} \right\rbrack^{2}}{\left\lbrack Ag^{+}(CN)_{2}^{-} \right\rbrack} \Rightarrow 1 \times 10^{- 20} = \frac{\left\lbrack Ag^{+} \right\rbrack \times \left( 2.5 \times 10^{- 7} \right)^{2}}{0.01} $$$\therefore\left\lbrack Ag^{+} \right\rbrack = 1.6 \times 10^{- 9}\text{ M}$$
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