Question

The formation constant* of [M(CN)6]4– is 2.50 × 1017, where M is a generic metal. A 0.170-mole quantity of M(NO3)2 is added to a liter of 1.41 M NaCN solution.

What is the concentration of M2 ions at equilibrium?

Answer 1

The reaction is M+2 ( from M(NO3)2 + 6CN- ( from NaCN) --------> [M(CN)6]-4

moles of M+2= 0.170, moles of NaCN= 1.41*1= 1.41

Moles of CN- that react= 0.170*6= 1.02 and moles of CN- remain =1.41-1.02=0.39 moles and moles of Complex= 0.17 moles

Concentrations : CN- = 0.39/1= 0.39M, [M(CN)6]^-4 =0.17

Formation constant Kf= [M(CN)6]-4 / [Mn+2] [CN-]6

= 2.5*10¹⁷= 0.17/ {[Mn+2] 0.39⁶}

Mn+2 =0.17/ *2.5*10¹⁷*0.00351 = 1.932*10-¹⁶M

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