Concentration of H2SO4 and Na2SO4 in a solution is 1 M and 1.8 x 10^-2 M, respectively. Molar solubi

Q: Why do we assume the first dissociation of $\text{H}_2\text{SO}_4$ is complete?

Because $K_{a1}$ for sulfuric acid is very large, meaning it acts as a strong acid for the first proton, dissociating almost 100% in aqueous solution.

Q: Why did the $\text{HSO}_4^-$ equilibrium shift backward?

The initial concentration of sulfate ions from $\text{Na}_2\text{SO}_4$ makes the reaction quotient $Q_c$ greater than $K_{a2}$. According to Le Chatelier's principle, the system shifts backward to reach equilibrium.

Q: How can we just ignore $x$ in $1+x$ and $1-x$?

Since $K_{a2}$ is small ($1.2 \times 10^{-2}$) and the equilibrium shifts backward, the change $x$ is very small compared to $1\text{ M}$. Approximating $1 \pm x \approx 1$ simplifies the math without losing significant accuracy.

Q: Why do we ignore $s$ in the total sulfate concentration for the $K_{sp}$ calculation?

Lead sulfate is a sparingly soluble salt with a very small $K_{sp}$ ($1.6 \times 10^{-8}$). The sulfate contributed by its dissolution ($s$) is negligible compared to the $1.2 \times 10^{-2}\text{ M}$ already present in the solution.

Q: What is the common ion effect?

It is the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. Here, the presence of sulfate from other sources suppresses the dissolution of lead sulfate.

Question

Concentration of $	ext{H}_2	ext{SO}_4$ and $	ext{Na}_2	ext{SO}_4$ in a solution is $1	ext{ M}$ and $1.8 	imes 10^{-2}	ext{ M}$, respectively. Molar solubility of $	ext{PbSO}_4$ in the same solution is $	ext{X} 	imes 10^{-	ext{Y}}	ext{ M}$ (expressed in scientific notation). The value of $	ext{Y}$ is _________.
[Given: Solubility product of $	ext{PbSO}_4$ ($K_{	ext{sp}}$) = $1.6 	imes 10^{-8}$. For $	ext{H}_2	ext{SO}_4$, $K_{	ext{a}1}$ is very large and $K_{	ext{a}2} = 1.2 	imes 10^{-2}$]

Sigma Percentile · Accepted Answer

Imagine you are a molecular detective, and you've just been handed a beaker containing a complex chemical soup. Inside this beaker, multiple reactions are happening simultaneously, each trying to reach its own state of balance. This is the beautiful, chaotic world of simultaneous equilibria.

In this problem, we are asked to find the solubility of lead sulfate, $	ext{PbSO}_4$, in a solution that already contains sulfuric acid, $	ext{H}_2	ext{SO}_4$, and sodium sulfate, $	ext{Na}_2	ext{SO}_4$. At first glance, it might seem overwhelming. But don't worry! We will break it down step by step, isolating each reaction and understanding how they interact.

### Analyzing the Setup

First, let's take inventory of what we have in our beaker before any weak equilibria establish themselves. We have strong electrolytes that dissociate completely.

Sulfuric acid is a strong acid for its first proton. Because $K_{a1}$ is very large, $1	ext{ M}$ of $	ext{H}_2	ext{SO}_4$ will completely dissociate to give $1	ext{ M}$ of $	ext{H}^+$ and $1	ext{ M}$ of $	ext{HSO}_4^-$.

Similarly, sodium sulfate is a highly soluble salt. $1.8 	imes 10^{-2}	ext{ M}$ of $	ext{Na}_2	ext{SO}_4$ will completely dissociate to give $3.6 	imes 10^{-2}	ext{ M}$ of $	ext{Na}^+$ and $1.8 	imes 10^{-2}	ext{ M}$ of $	ext{SO}_4^{2-}$.

So, our initial playing field is set. We have $	ext{H}^+$, $	ext{HSO}_4^-$, and $	ext{SO}_4^{2-}$ floating around. Now, the real chemistry begins.

### The Master Equation

The bisulfate ion, $	ext{HSO}_4^-$, is a weak acid. It wants to establish its own equilibrium:

$$ 	ext{HSO}_4^- ightleftharpoons 	ext{H}^+ + 	ext{SO}_4^{2-} $$

But wait! The products of this reaction are already present in the solution. To know which way this reaction will shift, we must calculate the reaction quotient, $Q_c$.

$$ Q_c = \frac{[	ext{H}^+][	ext{SO}_4^{2-}]}{[	ext{HSO}_4^-]} = \frac{1 	imes 1.8 	imes 10^{-2}}{1} = 1.8 	imes 10^{-2} $$

Comparing this to the given equilibrium constant, $K_{a2} = 1.2 	imes 10^{-2}$, we see that $Q_c > K_{a2}$. According to **Le Chatelier's Principle**, the system has too much product and will shift backward to reach equilibrium.

Let's assume an amount $x$ of $	ext{SO}_4^{2-}$ reacts with $	ext{H}^+$ to form $	ext{HSO}_4^-$. At equilibrium, our concentrations will be:
$[	ext{HSO}_4^-] = 1 + x$
$[	ext{H}^+] = 1 - x$
$[	ext{SO}_4^{2-}] = 1.8 	imes 10^{-2} - x$

Now, we plug these into the $K_{a2}$ expression:

$$ K_{a2} = \frac{(1-x)(1.8 	imes 10^{-2} - x)}{1+x} = 1.2 	imes 10^{-2} $$

Here is where we use our chemical intuition to simplify the math. Since $x$ is going to be very small compared to $1$, we can safely approximate $1 - x \approx 1$ and $1 + x \approx 1$.

This beautifully simplifies our equation to:

$$ 1.8 	imes 10^{-2} - x = 1.2 	imes 10^{-2} $$

Solving for $x$, we get $x = 0.6 	imes 10^{-2}	ext{ M}$.

Therefore, the final, stable concentration of sulfate ions in our beaker is:

$$ [	ext{SO}_4^{2-}] = 1.8 	imes 10^{-2} - 0.6 	imes 10^{-2} = 1.2 	imes 10^{-2}	ext{ M} $$

### Final Calculation

Now that the stage is perfectly set, we introduce our sparingly soluble salt, $	ext{PbSO}_4$. It tries to dissolve:

$$ 	ext{PbSO}_4(s) ightleftharpoons 	ext{Pb}^{2+} + 	ext{SO}_4^{2-} $$

Let its solubility be $s$. The concentration of $	ext{Pb}^{2+}$ will be $s$. But what about $	ext{SO}_4^{2-}$? It will be $s$ plus the $1.2 	imes 10^{-2}	ext{ M}$ already present. This is the classic **Common Ion Effect**.

The solubility product expression is:

$$ K_{sp} = [	ext{Pb}^{2+}][	ext{SO}_4^{2-}] = s(s + 1.2 	imes 10^{-2}) = 1.6 	imes 10^{-8} $$

Because $K_{sp}$ is incredibly small, $s$ will be minuscule. We can confidently approximate $s + 1.2 	imes 10^{-2} \approx 1.2 	imes 10^{-2}$.

$$ s(1.2 	imes 10^{-2}) = 1.6 	imes 10^{-8} $$

$$ s = \frac{1.6 	imes 10^{-8}}{1.2 	imes 10^{-2}} = \frac{1.6}{1.2} 	imes 10^{-6}	ext{ M} $$

The problem asks us to express this in the form $X 	imes 10^{-Y}$. By comparing our result, we can clearly see that the exponent $Y$ is exactly **6**.

And there we have it! By carefully untangling the simultaneous equilibria and making smart approximations, we've solved a complex system. Always remember to look at the big picture before diving into the math!

Enter Numerical Value:

Visualized Solution

$Initial Setup$

$Strong Electrolytes$

$Weak Acid Equilibrium$

$Reaction Direction$

$Equilibrium State$

$Approximation$

$Sulfate Concentration$

$Solubility of PbSO_{4}$

$Solving for s$

$Final Answer$

$Conclusion$

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; } Initial Setup

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solubility of PbSO4​

.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; } Final Answer

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