# Chemistry

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.150-mole quantity of M(NO3)2 is added to a liter of 0.870 M NaCN solution. What is the concentration of M2 ions at equilibrium?

1. 👍 0
2. 👎 0
3. 👁 162
1. This is a little hard to show in writing but here goes.
......M^2+ + 4CN^- ==> [M(CN)4]^2-
I..0.150....0.870.......0
C.-0.150...-0.600....+0.150
E.....0.....0.270.....0.150

The first equilibrium shows the formation of the complex on the right. With such a large Kf of over 10^16, it is obvious that essentially ALL of the M^2+ (from the M(NO3)2) reacts to form the complex.

Now that the complex is formed, we reverse the equilibrium (so it appears we are starting with [M(CN)4]^2- to form M^+ and CN^-. In this respect it's just like a Ka or Kb problem in which you're looking for the products. The E line for the first equilibrium is the I line for the reverse equilibrium. So it looks this way.
M^+ + 4CN^- ==> [M(CN)4]^2-
0....0.270......0.150.......I
x......4x........-x.........C
x..0.270+4x....0.150-x......E

Plug the E line into the Kf expression and solve for x = (M^2+).
Post your work if you get stuck and I can help you through it.

1. 👍 0
2. 👎 0
posted by DrBob222
2. Where can I find the Kf expression?

1. 👍 0
2. 👎 0
posted by Meghan
3. I don't understand, DrBob222.

1. 👍 0
2. 👎 0
posted by Meghan

## Similar Questions

1. ### chemistry

The formation constant* of [M(CN)4]2– is 7.70 × 1016, where M is a generic metal. A 0.170-mole quantity of M(NO3)2 is added to a liter of 1.040 M NaCN solution. What is the concentration of M2 ions at equilibrium?

asked by Emily on March 25, 2016
2. ### Chemistry

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 1.31 M NaCN solution. What is the concentration of M2 ions at equilibrium?

asked by Nat on May 6, 2014
3. ### chemistry

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 1.51 M NaCN solution. What is the concentration of M2 ions at equilibrium?

asked by kitor on July 13, 2013
4. ### Chemistry

The formation constant of [M(CN)4]^-2 is 7.70 × 10^16, where M is a generic metal. A 0.150-mole quantity of M(NO3)2 is added to a liter of 0.840 M NaCN solution. What is the concentration of M2 ions at equilibrium?

asked by shyamjp on February 1, 2015
5. ### Chemistry Problem

Consider an ionic compound, MX, composed of generic metal M and generic halogen X. The enthalpy of formation of MX is ÄHf° = –457 kJ/mol. The enthalpy of sublimation of M is ÄHsub = 121 kJ/mol. The ionization energy of M is

asked by Anonymous on February 2, 2013
6. ### Chemistry

Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX2 is ΔHf° = –799 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 157 kJ/mol. The first and second

asked by Joseoh on October 30, 2016
7. ### Chemistry

At 25 °C, an aqueous solution has an equilibrium concentration of 0.00147 M for a generic cation, A (aq), and 0.000735 M for a generic anion, B2–(aq). What is the equilibrium constant, Ksp, of the generic salt A2B(s)?

asked by Jordan on May 6, 2016
8. ### Chemistry

Consider an ionic compound, MX2, composed of generic metal M and generic, gaseous halogen X. The enthalpy of formation of MX2 is ΔHf° = –799 kJ/mol. The enthalpy of sublimation of M is ΔHsub = 157 kJ/mol. The first and second

asked by Joseph on October 31, 2016
9. ### CHEM

3. which of the following is not likely to indicate that a double-displacement reaction has occurred neutralization of acids and bases, formation of a precipitate, formation of a gas, change of a metal ion to a metal element

asked by Anonymous on December 19, 2011
10. ### Chemistry

Consider an ionic compound, MX2, composed of generic metal M and generic halogen X. The enthalphy of formation of MX2= -891. The enthalphy of sublimation of M= 141. The first and second ionization energies of M are 605 and 1392.

asked by Garcia on November 29, 2012

More Similar Questions