Problem: When 0.84 g of ZnCl2 is dissolved in 245 mL of 0.150 M NaCN, what are [Zn2+], [Zn(CN)4 2−], and [CN−] [Kf of Zn(CN)4 2− = 4.2×1019]?

FREE Expert Solution

We’re being asked to determine [Zn2+], [Zn(CN)42−], and [CN] when 0.84 g of ZnCl2 is dissolved in 245 mL of 0.150 M NaCN

Since the compounds are ionic compounds, they form ions when dissociating in water

The dissociation of ZnCl2 and NaCN in water are as follows:

ZnCl2  Zn2+(aq) + 2 Cl(aq)

NaCN  Na+(aq) + CN(aq)

When Zn2+ and CN are mixed, the complex [Zn(CN)42–] is formed:

Zn2+(aq) + 4 CN(aq)  Zn(CN)42–

The equilibrium constant for this is called the formation constant (Kf) and is associated with the formation of a compound from two ions. Kf is also the opposite of Ksp.

The Kf expression for this is:

Note that each concentration is raised by the stoichiometric coefficient:

[Zn2+] and [Zn(CN)42–] are raised to 1 while [CN] is raised to 4.

We need to do the following for this problem:

Step 1: Calculate the molarity of Zn2+ and CN.
Step 2: Construct an ICE table for the reaction.
Step 3: Calculate the equilibrium concentrations.

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Problem Details

When 0.84 g of ZnCl2 is dissolved in 245 mL of 0.150 M NaCN, what are [Zn2+], [Zn(CN)4 2−], and [CN] [Kf of Zn(CN)4 2− = 4.2×1019]?