The chemistry of silver cyanide: (a) Calculate the solubility of AgCN(s) in water from the Ksp

The chemistry of silver cyanide: (a) Calculate the...

The chemistry of silver cyanide: (a) Calculate the solubility of $\mathrm{AgCN}(\mathrm{s})$ in water from the $K_{\text {sp }}$ value. (b) Calculate the value of the equilibrium constant for the reaction $$ \mathrm{AgCN}(\mathrm{s})+\mathrm{CN}^{-}(\mathrm{aq}) \rightleftarrows\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}(\mathrm{aq}) $$ from $K_{\mathrm{sp}}$ and $K_{\mathrm{f}}$ values and predict from this value whether $\mathrm{AgCN}(\mathrm{s})$ would dissolve in KCN(aq). (c) Determine the equilibrium constant for the reaction $$ \begin{aligned} & \mathrm{AgCN}(\mathrm{s})+2 \mathrm{~S}_2 \mathrm{O}_3{ }^{2-}(\mathrm{aq}) \rightleftarrows \\ & {\left[\mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3\right)_2\right]^{3-}(\mathrm{aq})+\mathrm{CN}^{-}(\mathrm{aq})} \\ & \end{aligned} $$ Calculate the solubility of $\mathrm{AgCN}$ in a solution containing $0.10 \mathrm{M} \mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ and compare the value to the solubility in water [part (a)].

The chemistry of silver cyanide:
(a) Calculate the solubility of $\mathrm{AgCN}(\mathrm{s})$ in water from the $K_{\text {sp }}$ value.
(b) Calculate the value of the equilibrium constant for the reaction
$$
\mathrm{AgCN}(\mathrm{s})+\mathrm{CN}^{-}(\mathrm{aq}) \rightleftarrows\left[\mathrm{Ag}(\mathrm{CN})_2\right]^{-}(\mathrm{aq})
$$
from $K_{\mathrm{sp}}$ and $K_{\mathrm{f}}$ values and predict from this value whether $\mathrm{AgCN}(\mathrm{s})$ would dissolve in KCN(aq).
(c) Determine the equilibrium constant for the reaction
$$
\begin{aligned}
& \mathrm{AgCN}(\mathrm{s})+2 \mathrm{~S}_2 \mathrm{O}_3{ }^{2-}(\mathrm{aq}) \rightleftarrows \\
& {\left[\mathrm{Ag}\left(\mathrm{S}_2 \mathrm{O}_3\right)_2\right]^{3-}(\mathrm{aq})+\mathrm{CN}^{-}(\mathrm{aq})} \\
&
\end{aligned}
$$

Calculate the solubility of $\mathrm{AgCN}$ in a solution containing $0.10 \mathrm{M} \mathrm{S}_2 \mathrm{O}_3{ }^{2-}$ and compare the value to the solubility in water [part (a)].

Key Concepts

Solubility Product (Ksp)

The solubility product constant (Ksp) represents the equilibrium between a sparingly soluble salt and its ions in solution. It quantifies the maximum product of the molar concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient, at saturation. Understanding and applying Ksp allows one to calculate the solubility of a solid in water and predict precipitation conditions.

Complex Ion Formation

Complex ion formation occurs when a metal ion binds with one or more ligands to form a complex soluble species. This process alters the solubility of the original salt by shifting the equilibrium, often increasing solubility. The formation of complex ions is a key concept in coordination chemistry and is used to control and predict solubility and reaction outcomes.

Formation Constant (Kf)

The formation constant (Kf) is a specific equilibrium constant that quantifies the stability of a complex ion formed in solution. A larger Kf indicates a more stable complex, meaning the equilibrium strongly favors complex formation. It is essential in predicting how ligands will interact with metal ions to alter solubility equilibria by drawing the metal ion from other equilibria, such as precipitation.

Equilibrium Constant Manipulation

Many chemical reactions can be understood as the combination of simpler equilibrium steps. By manipulating and combining equilibrium constants (such as Ksp and Kf), one can derive the overall equilibrium constant for a complex reaction. This process is fundamental in predicting reaction direction and the extent of reaction under given conditions.

Common Ion Effect and Ligand Exchange

The common ion effect refers to the shift in equilibrium caused by the addition of an ion that is already a component of the equilibrium system, typically reducing the solubility of a salt. In conjunction with ligand exchange, where one ligand is replaced by another in a complex, these effects can significantly impact solubility and overall reaction equilibrium. They are pivotal in understanding how competing reactions and the presence of additional ions affect chemical equilibria.

Transcript

00:01 This one, as you guessed, is a fairly lengthy problem with multiple parts.

00:05 First, let's write the equilibrium that corresponds to ksp for silver cyanide.

00:11 Silver cyanide solid goes to silver ion plus cyanide anion.

00:18 It has a ksp equal to the product of the silver and cyanide concentrations, which could be expressed as molar solubility squared.

00:27 And when we look up the value, we get 6 .0 times 10 to the negative 17.

00:32 So molar solubility in the presence of nothing else but pure water is going to be the square root of 6 .0 times 10 to negative 17 or 7 .75 10 to negative 9 molar.

00:46 Now moving on to part b, we're asked to calculate the value of the equilibrium constant for the reaction that's given to us.

00:53 If we take, which is this reaction here, if we take the silver cyanide, going to silver ion and cyanide ion, that is ksp, that is the ksp equilibrium.

01:07 If we then look at the kf value for silver plus two cyanides going to the complex ion, if we sum these two reactions together, we get the above reaction.

01:22 So if we sum these two reactions, then the equilibrium constant for the above reaction is going to be the product of the equilibrium constant for these two reactions.

01:32 The first one is 6 .0 times 10 to negative 17.

01:35 The second one we look up for this reaction is 1 .3, 10 to the 21.

01:40 So we get 7 .8 times 10 to the 4...

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