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The formation constant* of [M(CN)2]– is 5.30 × 1018, where M is a generic metal. A...

The formation constant* of [M(CN)2]– is 5.30 × 1018, where M is a generic metal. A 0.170-mole quantity of M(NO3) is added to a liter of 0.540 M NaCN solution. What is the concentration of M ions at equilibrium?

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Answer #1

MNO3(aq) + 2NaCN <------> [M(CN)2]-(aq) ; Kf = 5.3*1018

Kf value being too large indicates that all the M+ ions are used up

Thus, CN- left = 0.54 - 2*0.17 = 0.2 moles

Now, 5.3*1018 = [M(CN)2-]/{[MNO3]*[NaCN]2}

or, 5.3*1018 = 0.17/(0.22*[MNO3]}

or, [MNO3] = 8.02*10-19 M = [M+]

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