A solution is 0.1 M in Cl- and 0.001 M in CrO4(2-). Solid AgNO3 is gradually added to it. Assuming t

Q: Why can't we just say Ag2CrO4 precipitates first because its Ksp is lower ($10^{-12}$ vs $10^{-10}$)?

You cannot directly compare $K_{sp}$ values to determine solubility or precipitation order if the salts have different stoichiometries (like 1:1 for AgCl and 2:1 for Ag2CrO4). You must calculate the actual concentration of the common ion needed.

Q: Why do we use the initial concentrations of Cl- and CrO4(2-) in the Ksp expression?

We are calculating the exact moment precipitation *begins*. At this threshold, the concentrations of the anions haven't decreased yet because no precipitate has formed. So, we use their initial concentrations.

Q: What happens to the volume when we add solid AgNO3?

The problem explicitly states to assume that the addition of solid $\text{AgNO}_3$ does not change the volume of the solution. This allows us to treat the concentrations of $\text{Cl}^-$ and $\text{CrO}_4^{2-}$ as constant until precipitation starts.

Q: Why did we write $1.9 \times 10^{-9}$ as $19 \times 10^{-10}$?

To easily calculate the square root without a calculator. By making the exponent an even number ($-10$), we can simply halve it to $-5$ when taking the square root, and $\sqrt{19}$ is roughly between 4 and 5 (specifically $\approx 4.3$).

Q: Will Ag2CrO4 ever precipitate?

Yes! As you continue to add $\text{AgNO}_3$, $\text{AgCl}$ will keep precipitating, and the $[\text{Ag}^+]$ in solution will slowly rise. Once the $[\text{Ag}^+]$ reaches $4.3 \times 10^{-5}\text{ M}$, $\text{Ag}_2\text{CrO}_4$ will also start to precipitate alongside $\text{AgCl}$.

Question

A solution is $0.1	ext{ M}$ in $	ext{Cl}^-$ and $0.001	ext{ M}$ in $	ext{CrO}_4^{2-}$. Solid $	ext{AgNO}_3$ is gradually added to it.
Assuming that the addition does not change in volume and $K_{sp}(	ext{AgCl}) = 1.7 	imes 10^{-10}	ext{ M}^2$ and $K_{sp}(	ext{Ag}_2	ext{CrO}_4) = 1.9 	imes 10^{-12}	ext{ M}^3$
Select correct statement from the following

Options:

(A) $	ext{AgCl}$ precipitates first because its $K_{sp}$ is high.

(B) $	ext{Ag}_2	ext{CrO}_4$ precipitates first as its $K_{sp}$ is low.

(C) $	ext{Ag}_2	ext{CrO}_4$ precipitates first because the amount of $	ext{Ag}^+$ needed is low.

(D) $	ext{AgCl}$ will precipitate first as the amount of $	ext{Ag}^+$ needed to precipitate is low.

Sigma Percentile · Accepted Answer

### The Battle of the Precipitates

Imagine you are a referee in a microscopic race. In a single beaker, you have two different anions: chloride ($	ext{Cl}^-$) and chromate ($	ext{CrO}_4^{2-}$), swimming around peacefully.

Suddenly, you start dropping in silver ions ($	ext{Ag}^+$). Both anions want to bond with the silver to form a solid precipitate. But who will win? Who will precipitate first?

This is a classic problem of **fractional precipitation**. To determine the winner, we cannot simply look at who has the lower solubility product ($K_{sp}$).

### The Trap of Comparing $K_{sp}$ Directly

A very common silly mistake is to look at the $K_{sp}$ values and say, "Ah! $1.9 	imes 10^{-12}$ is smaller than $1.7 	imes 10^{-10}$, so silver chromate must precipitate first!"

**This is completely wrong.**

Why? Because the two salts have different stoichiometries. Silver chloride ($	ext{AgCl}$) is a 1:1 salt, meaning it needs one silver ion for every chloride ion. Silver chromate ($	ext{Ag}_2	ext{CrO}_4$) is a 2:1 salt, requiring *two* silver ions for every chromate ion.

Because of this difference, their $K_{sp}$ expressions scale differently with the concentration of silver ions. The only mathematically rigorous way to find the winner is to calculate the exact threshold concentration of $	ext{Ag}^+$ needed to trigger precipitation for each salt individually.

### The Mathematical Threshold

Let's calculate the threshold for **Silver Chloride ($	ext{AgCl}$)** first. Precipitation begins the exact moment the ionic product equals the $K_{sp}$.

$$ K_{sp}(	ext{AgCl}) = [	ext{Ag}^+][	ext{Cl}^-] $$

Substituting the given values:

$$ 1.7 	imes 10^{-10} = [	ext{Ag}^+](0.1) $$

Solving for the silver ion concentration gives us:

$$ [	ext{Ag}^+] = 1.7 	imes 10^{-9} 	ext{ M} $$

This is the minimum amount of silver needed to start forming white $	ext{AgCl}$ precipitate.

Now, let's do the same for **Silver Chromate ($	ext{Ag}_2	ext{CrO}_4$)**. Its $K_{sp}$ expression involves a squared term because of the two silver ions in its formula.

$$ K_{sp}(	ext{Ag}_2	ext{CrO}_4) = [	ext{Ag}^+]^2[	ext{CrO}_4^{2-}] $$

Substituting the given values:

$$ 1.9 	imes 10^{-12} = [	ext{Ag}^+]^2(0.001) $$

$$ [	ext{Ag}^+]^2 = 1.9 	imes 10^{-9} = 19 	imes 10^{-10} $$

Taking the square root:

$$ [	ext{Ag}^+] = \sqrt{19} 	imes 10^{-5} \approx 4.3 	imes 10^{-5} 	ext{ M} $$

### The Verdict

We now have our two thresholds.

To precipitate $	ext{AgCl}$, we need $[	ext{Ag}^+] = 1.7 	imes 10^{-9} 	ext{ M}$.
To precipitate $	ext{Ag}_2	ext{CrO}_4$, we need $[	ext{Ag}^+] = 4.3 	imes 10^{-5} 	ext{ M}$.

As we slowly add $	ext{AgNO}_3$ to the beaker, the concentration of $	ext{Ag}^+$ rises from zero. It will hit the smaller number first.

Since $1.7 	imes 10^{-9}$ is significantly smaller than $4.3 	imes 10^{-5}$, the threshold for silver chloride is reached much earlier. Therefore, **$	ext{AgCl}$ will precipitate first**, simply because the amount of $	ext{Ag}^+$ needed to precipitate it is lower.

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Visualized Solution

$Visualizing the Setup$

$Condition for Precipitation$

$Setup for AgCl$

$Computing [Ag^{+}] for AgCl$

$Setup for Ag_{2} CrO_{4}$

$Computing [Ag^{+}] for Ag_{2} CrO_{4}$

$Conclusion$

The Sigma Insight: Solubility Product and Common Ion Effect

The Battle of the Precipitates

The Trap of Comparing $K_{s p}$ Directly

The Mathematical Threshold

The Verdict

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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; } Condition for Precipitation

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Setup for Ag2​CrO4​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Computing [Ag+] for Ag2​CrO4​

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