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Question: Solubility of \(NH_{3}\) is \(NH_{3}\)mol \(I_{2}\)The value of \(I_{2(g)}\) for the salt is....

Solubility of NH3NH_{3} is NH3NH_{3}mol I2I_{2}The value of I2(g)I_{2(g)} for the salt is.

A

2I(g);ΔfHo=+150kJ2I_{(g)};\Delta_{f}H^{o} = + 150kJ

B

2.5×10−162.5 \times 10 ^ { - 16 }

C

[Fe(SCN)]Deepred(aq)2+;{\underset{Deepred}{\left\lbrack Fe(SCN) \right\rbrack}}_{(aq)}^{2 +};

D

Kc=[Fe(SCN)]2+[Fe3+][SCN−]K_{c} = \frac{\left\lbrack Fe(SCN) \right\rbrack^{2 +}}{\left\lbrack Fe^{3 +} \right\rbrack\left\lbrack SCN^{-} \right\rbrack}

Answer

Kc=[Fe(SCN)]2+[Fe3+][SCN−]K_{c} = \frac{\left\lbrack Fe(SCN) \right\rbrack^{2 +}}{\left\lbrack Fe^{3 +} \right\rbrack\left\lbrack SCN^{-} \right\rbrack}

Explanation

Solution

: Ca2+0.5×10−4+2F−(2×0.5×10=4)2\underset{0.5 \times 10^{- 4}}{Ca^{2 +}} + \underset{(2 \times 0.5 \times 10^{= 4})^{2}}{2F^{-}}

Ksp=0.5×10−4×1×10−8=0.5×10−12K_{sp} = 0.5 \times 10^{- 4} \times 1 \times 10^{- 8} = 0.5 \times 10^{- 12} or 5×10−135 \times 10^{- 13}