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Question: The degree of ionisation of an acid HA is \(K_{c}\)at \(2Cu(NO_{3})_{2(s)}\)M concentratin. Its. Dis...

The degree of ionisation of an acid HA is KcK_{c}at 2Cu(NO3)2(s)2Cu(NO_{3})_{2(s)}M concentratin. Its. Dissociation constant will be

A

Kc=[NO2(g)]4[O2(g)][Cu(NO3)2(s)]2K_{c} = \frac{\left\lbrack NO_{2(g)} \right\rbrack^{4}\left\lbrack O_{2(g)} \right\rbrack}{\left\lbrack Cu(NO_{3})_{2(s)} \right\rbrack^{2}}

B

Kc=[NO2(g)]4[O2(g)]K_{c} = \left\lbrack NO_{2(g)} \right\rbrack^{4}\left\lbrack O_{2(g)} \right\rbrack

C

Kc=[CuO(s)]2[Cu(NO3)2(s)]2K_{c} = \frac{\left\lbrack CuO_{(s)} \right\rbrack^{2}}{\left\lbrack Cu(NO_{3})_{2(s)} \right\rbrack^{2}}

D

KcK_{c}

Answer

Kc=[NO2(g)]4[O2(g)]K_{c} = \left\lbrack NO_{2(g)} \right\rbrack^{4}\left\lbrack O_{2(g)} \right\rbrack

Explanation

Solution

: Ka=cα2=0.1×(10−5)2=10−11K_{a} = c\alpha^{2} = 0.1 \times \left( 10^{- 5} \right)^{2} = 10^{- 11}