The solubility product of PbI2 is 8.0 x 10^-9. The solubility of lead iodide in 0.1 molar solution o

Q: Why do we ignore $S$ in the expression $0.1 + S$?

Because $K_{sp}$ is extremely small ($8.0 \times 10^{-9}$), the solubility $S$ will be tiny (around $10^{-4}$). Adding $0.0001$ to $0.1$ gives $0.1001$, which is practically just $0.1$. Ignoring $S$ makes the math much simpler without losing significant accuracy.

Q: Why is $[\text{I}^-]$ taken as $2S$ and then squared in the formula?

The concentration of iodide ions is $2S$ because each molecule of $\text{PbI}_2$ releases two $\text{I}^-$ ions. The formula for $K_{sp}$ requires us to raise the concentration of each ion to the power of its stoichiometric coefficient. So, we must square the entire concentration $(2S)$, resulting in $4S^2$.

Q: Does the volume of the solution matter here?

No, because we are already given the concentration in molarity ($\text{mol/L}$). The solubility product $K_{sp}$ and solubility $S$ are intensive properties that depend on concentration, not the total volume of the solution.

Q: What if the approximation $S \ll 0.1$ is not valid?

If $K_{sp}$ were much larger, $S$ would be significant. In that case, you couldn't ignore it, and you would have to solve the full cubic equation $(0.1 + S)(2S)^2 = K_{sp}$. Thankfully, for most JEE problems, the approximation holds true!

Q: Why did we shift the decimal at the end?

The question specifically asked for the answer in the format $x \times 10^{-6}$. Our calculated solubility was $1.41 \times 10^{-4}$. To match the requested power of $10^{-6}$, we multiply the coefficient by $100$ and divide the power by $100$, giving $141 \times 10^{-6}$.

Question

The solubility product of $	ext{PbI}_2$ is $8.0 	imes 10^{-9}$. The solubility of lead iodide in $0.1$ molar solution of lead nitrate is $x 	imes 10^{-6} 	ext{ mol/L}$. The value of $x$ is ....... (Rounded off to the nearest integer).
[Given, $\sqrt{2} = 1.41$]

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a beaker filled with a $0.1	ext{ M}$ solution of lead nitrate, $	ext{Pb(NO}_3)_2$. Into this solution, we drop some solid lead iodide, $	ext{PbI}_2$. We want to find out exactly how much of this lead iodide will dissolve.

This is a classic scenario of the **Common Ion Effect**. Lead nitrate is a strong electrolyte, meaning it completely dissociates in water to give a massive influx of $	ext{Pb}^{2+}$ and $	ext{NO}_3^-$ ions. On the other hand, lead iodide is sparingly soluble and sets up a delicate equilibrium:

$	ext{PbI}_2(s) ightleftharpoons 	ext{Pb}^{2+}(aq) + 2	ext{I}^-(aq)$

Because the solution is already flooded with $	ext{Pb}^{2+}$ ions from the lead nitrate, Le Chatelier's principle tells us that the equilibrium of lead iodide will be pushed heavily to the left. Its solubility will be drastically reduced!

### The Master Equation

Let the solubility of $	ext{PbI}_2$ in this solution be $S$. The total concentration of lead ions in the solution comes from both sources:

$[	ext{Pb}^{2+}]_{	ext{total}} = 0.1	ext{ M (from nitrate)} + S	ext{ (from iodide)}$

Since $S$ is going to be incredibly small compared to $0.1$, we can safely approximate the total lead ion concentration to just $0.1	ext{ M}$. The concentration of iodide ions, however, comes entirely from the dissolving $	ext{PbI}_2$, so $[	ext{I}^-] = 2S$.

Now, we bring in our master tool, the solubility product constant ($K_{sp}$):

$K_{sp} = [	ext{Pb}^{2+}][	ext{I}^-]^2$

Substituting our known values into this raw structure:

$8.0 	imes 10^{-9} = (0.1)(2S)^2$

### Final Calculation

Let's carefully execute the algebra. Don't make a silly mistake with the square!

$8.0 	imes 10^{-9} = 0.1 	imes 4S^2$

Dividing both sides by $0.1$ (which is equivalent to multiplying by $10$) gives:

$4S^2 = 8.0 	imes 10^{-8}$

Now, divide by $4$:

$S^2 = 2.0 	imes 10^{-8}$

Taking the square root of both sides is straightforward. We are given that $\sqrt{2} = 1.41$:

$S = \sqrt{2} 	imes 10^{-4} = 1.41 	imes 10^{-4}	ext{ M}$

We have our solubility! But wait, there is a catch. The question specifically asks for the answer in the format of $x 	imes 10^{-6}$. To match this, we shift the decimal point two places to the right, which decreases the exponent by two:

$S = 141 	imes 10^{-6}	ext{ M}$

Thus, our integer **$x$ is exactly $141$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Setup$

$Dissociation Equations$

$The Common Ion Effect$

$Setting up K_{s p}$

$Algebraic Manipulation$

$Solving for S$

$Formatting the Answer$

$The Way Forward$

The Sigma Insight: Solubility Product and Common Ion Effect

Analyzing the Setup

The Master Equation

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 Setup

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

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

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

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

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

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

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