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?
MNO_{3}(aq) + 2NaCN <------> [M(CN)_{2}]^{-}(aq) ; K_{f} = 5.3*10^{18}
K_{f} 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*10^{18} = [M(CN)_{2}^{-}]/{[MNO_{3}]*[NaCN]^{2}}
or, 5.3*10^{18} = 0.17/(0.2^{2}*[MNO_{3}]}
or, [MNO_{3}] = 8.02*10^{-19} M = [M^{+}]
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 x 1018, where M is a generic metal. A 0.170 mole quantity of M(NO3) is added to a liter of 0.650 M NaCN solution. What is the concentration of M+ ions at equilibrium? [M2+] =
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can someone please help me answer these, for some reason theh are both wrong and i cant get them right, will rate you good if you can help get both correct answers! thanks The formation constant of M(CN)61 is 2.50 x 10'7, where M is a generic metal. A 0.130 mole quantity of M(NO3) is added to a liter of 1.180 M NaCN solution. What is the concentration of M2+ ions at equilibrium? [M ] = 4.6 x10-19 The generic...
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