An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2S. What is the minimum molar concentr

Q: Why do we equate the ionic product to $K_{\text{sp}}$ instead of making it less than $K_{\text{sp}}$?

Equating the ionic product to $K_{\text{sp}}$ gives us the boundary condition—the exact point where the solution is saturated but precipitation hasn't started. This allows us to calculate the *minimum* $\text{H}^+$ required. Any concentration of $\text{H}^+$ greater than this will make the ionic product less than $K_{\text{sp}}$, which is even safer.

Q: Why is the concentration of $\text{H}_2\text{S}$ taken as $0.1\text{ M}$ even after it dissociates?

$\text{H}_2\text{S}$ is a very weak acid, and its dissociation is further suppressed by the highly acidic medium (common ion effect from the added $\text{H}^+$). Therefore, the amount that dissociates is negligible compared to the initial $0.1\text{ M}$, so we can safely assume its equilibrium concentration remains $0.1\text{ M}$.

Q: Why do we use $K_{\text{NET}}$ instead of $K_1$ and $K_2$ separately?

Since we are dealing with the overall equilibrium between the initial reactant ($\text{H}_2\text{S}$) and the final products ($\text{H}^+$ and $\text{S}^{2-}$), using the overall dissociation constant $K_{\text{NET}} = K_1 \times K_2$ simplifies the calculation significantly. It directly relates the species we care about.

Q: What is the physical meaning of adding $\text{H}^+$ to prevent precipitation?

This is the common ion effect. By adding $\text{H}^+$ (an acid), you are adding a product of the $\text{H}_2\text{S}$ dissociation reaction. According to Le Chatelier's principle, this pushes the equilibrium to the left, reducing the amount of $\text{S}^{2-}$ produced. Less $\text{S}^{2-}$ means the ionic product of $\text{ZnS}$ stays below its $K_{\text{sp}}$.

Q: What would happen if the calculated $[\text{H}^+]$ was higher than what we can practically achieve?

If the required $[\text{H}^+]$ was extremely high (e.g., $>10\text{ M}$), it would mean that $\text{ZnS}$ is so insoluble that it's practically impossible to prevent its precipitation just by acidifying the solution. In such cases, complexing agents might be needed.

Question

An acidified solution of $0.05	ext{ M}$ $	ext{Zn}^{2+}$ is saturated with $0.1	ext{ M}$ $	ext{H}_2	ext{S}$. What is the minimum molar concentration (M) of $	ext{H}^+$ required to prevent the precipitation of $	ext{ZnS}$ ?
[Use $K_{	ext{sp}} (	ext{ZnS}) = 1.25 	imes 10^{-22}$ and Overall dissociation constant of $	ext{H}_2	ext{S}$ , $K_{	ext{NET}} = K_1K_2 = 1 	imes 10^{-21}$]

Sigma Percentile · Accepted Answer

The problem of preventing precipitation in an ionic solution is a classic application of chemical equilibrium and the common ion effect. Let's dive into the fascinating interplay of ions in this solution.

### Analyzing the Setup
Imagine a beaker containing a solution of zinc ions ($	ext{Zn}^{2+}$) at a concentration of $0.05	ext{ M}$. We saturate this solution with hydrogen sulfide ($	ext{H}_2	ext{S}$) gas, maintaining its concentration at $0.1	ext{ M}$.

The danger here is the formation of zinc sulfide ($	ext{ZnS}$), a sparingly soluble salt that tends to precipitate out. Our goal is to find the exact amount of acid (specifically, $	ext{H}^+$ ions) needed to keep the $	ext{ZnS}$ dissolved.

### The Precipitation Boundary
For any sparingly soluble salt, precipitation begins the moment the ionic product exceeds the solubility product ($K_{	ext{sp}}$). To *just* prevent precipitation, we must hold the system exactly at the brink of saturation.

This gives us our master boundary condition:
$$[	ext{Zn}^{2+}][	ext{S}^{2-}] = K_{	ext{sp}}$$

We know the concentration of zinc ions and the $K_{	ext{sp}}$ of zinc sulfide. Let's substitute these values to find the maximum allowable concentration of sulfide ions ($	ext{S}^{2-}$):
$$0.05 	imes [	ext{S}^{2-}] = 1.25 	imes 10^{-22}$$

Solving for $[	ext{S}^{2-}]$, we get:
$$[	ext{S}^{2-}] = \frac{1.25 	imes 10^{-22}}{0.05} = 2.5 	imes 10^{-21}	ext{ M}$$

This incredibly tiny number is the absolute maximum concentration of sulfide ions our solution can tolerate before $	ext{ZnS}$ starts crashing out as a solid.

### The Source of Sulfide: $	ext{H}_2	ext{S}$ Dissociation
Where do these sulfide ions come from? They are produced by the dissociation of the dissolved $	ext{H}_2	ext{S}$ gas. Hydrogen sulfide is a weak diprotic acid, and its overall dissociation can be represented as:
$$	ext{H}_2	ext{S} ightleftharpoons 2	ext{H}^+ + 	ext{S}^{2-}$$

The equilibrium constant for this overall reaction is given as $K_{	ext{NET}}$. The expression for this constant is:
$$K_{	ext{NET}} = \frac{[	ext{H}^+]^2[	ext{S}^{2-}]}{[	ext{H}_2	ext{S}]}$$

### The Common Ion Effect in Action
This is where the magic of the common ion effect comes into play. By adding an external source of $	ext{H}^+$ (acidifying the solution), we can push the $	ext{H}_2	ext{S}$ equilibrium to the left, according to Le Chatelier's principle. This suppresses the formation of $	ext{S}^{2-}$, keeping its concentration below our critical threshold.

Let's plug our known values into the $K_{	ext{NET}}$ expression. We know $K_{	ext{NET}} = 10^{-21}$, $[	ext{H}_2	ext{S}] = 0.1	ext{ M}$, and our maximum allowed $[	ext{S}^{2-}] = 2.5 	imes 10^{-21}	ext{ M}$:
$$10^{-21} = \frac{[	ext{H}^+]^2 	imes (2.5 	imes 10^{-21})}{0.1}$$

Now, we solve for $[	ext{H}^+]^2$:
$$[	ext{H}^+]^2 = \frac{10^{-21} 	imes 0.1}{2.5 	imes 10^{-21}}$$

Notice how beautifully the $10^{-21}$ terms cancel out:
$$[	ext{H}^+]^2 = \frac{0.1}{2.5} = \frac{1}{25}$$

### The Final Calculation
To find the minimum concentration of $	ext{H}^+$ required, we simply take the square root of both sides:
$$[	ext{H}^+] = \sqrt{\frac{1}{25}} = \frac{1}{5} = 0.2	ext{ M}$$

**Conclusion:** We must maintain a minimum $	ext{H}^+$ concentration of **$0.2	ext{ M}$** to suppress the dissociation of $	ext{H}_2	ext{S}$ just enough so that the sulfide ion concentration stays at or below $2.5 	imes 10^{-21}	ext{ M}$, thereby perfectly preventing the precipitation of zinc sulfide.

Enter Numerical Value:

Visualized Solution

$Visualizing the Solution$

$Condition for Precipitation$

$Substituting Values$

$Maximum Allowed [S^{2 -}]$

$Dissociation of H_{2} S$

$Setting up the Equilibrium Equation$

$Solving for [H^{+}]^{2}$

$Final Calculation$

$The Common Ion Effect$

The Sigma Insight: Solubility Product and Common Ion Effect

Analyzing the Setup Imagine a beaker containing a solution of zinc ions ( $Zn^{2 +}$ ) at a concentration of $0.05 M$

The Precipitation Boundary For any sparingly soluble salt, precipitation begins the moment the ionic product exceeds the solubility product ( $K_{sp}$ )

The Source of Sulfide: $H_{2} S$ Dissociation Where do these sulfide ions come from? They are produced by the dissociation of the dissolved $H_{2} S$ gas

The Common Ion Effect in Action This is where the magic of the common ion effect comes into play

The Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Precipitation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Maximum Allowed [S2−]

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Dissociation of H2​S

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setting up the Equilibrium Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for [H+]2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Common Ion Effect