In a saturated solution of the sparingly soluble strong electrolyte AgIO3 (molecular mass = 283) the

Q: Why do we ignore the solid $\text{AgIO}_3$ in the $K_{sp}$ expression?

The concentration of a pure solid is constant. Because it doesn't change during the reaction, it is mathematically incorporated into the equilibrium constant $K_{sp}$ itself, simplifying our calculations.

Q: What exactly does 'sparingly soluble' mean?

It means the substance dissolves to a very small extent. The concentration of its ions in solution is extremely low, which makes the approximations used in $K_{sp}$ calculations highly accurate.

Q: Why did we divide by 10 to find the moles in 100 mL?

Molarity ($S$) is defined as moles per $1000\text{ mL}$ ($1\text{ L}$). To find the moles in $100\text{ mL}$, which is exactly $1/10$th of a liter, we simply divide the molarity by 10.

Q: Can we use this $K_{sp}$ method for highly soluble salts like NaCl?

No, for highly soluble salts, the concept of $K_{sp}$ is not typically used. At high concentrations, ionic interactions become significant, and we must use 'activities' instead of simple concentrations.

Q: What would happen to the solubility if the temperature changes?

The solubility product constant, $K_{sp}$, is strictly temperature-dependent. If the temperature changes, the $K_{sp}$ value will change, which in turn will either increase or decrease the solubility of the salt.

Question

In a saturated solution of the sparingly soluble strong electrolyte $	ext{AgIO}_3$ (molecular mass $= 283$) the equilibrium which sets in is
$	ext{AgIO}_3(s) ightleftharpoons 	ext{Ag}^+(aq) + 	ext{IO}_3^-(aq)$
If the solubility product constant, $K_{sp}$ of $	ext{AgIO}_3$ at a given temperature is $1.0 	imes 10^{-8}$, what is the mass of $	ext{AgIO}_3$ contained in $100	ext{ mL}$ of its saturated solution?

Options:

(A) $28.3 	imes 10^{-2}	ext{ g}$

(B) $2.83 	imes 10^{-3}	ext{ g}$

(C) $1.0 	imes 10^{-7}	ext{ g}$

(D) $1.0 	imes 10^{-4}	ext{ g}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a beaker filled with water. When we add solid silver iodate ($	ext{AgIO}_3$) to it, it doesn't dissolve completely like table salt. Instead, it's a **sparingly soluble** salt. This means only a tiny fraction of it breaks apart into ions, while the rest remains as a solid at the bottom of the beaker.

This creates a dynamic equilibrium between the undissolved solid and the dissolved ions in the solution. We can represent this chemical dance with the following equation:

$$	ext{AgIO}_3(s) ightleftharpoons 	ext{Ag}^+(aq) + 	ext{IO}_3^-(aq)$$

### The Master Equation: Solubility Product
To quantify this equilibrium, we use the **solubility product constant**, denoted as $K_{sp}$. The $K_{sp}$ is simply the product of the molar concentrations of the dissolved ions, each raised to the power of their stoichiometric coefficients. Since both ions have a coefficient of 1, our expression is beautifully simple:

$$K_{sp} = [	ext{Ag}^+][	ext{IO}_3^-]$$

Notice that the solid $	ext{AgIO}_3$ is not included in this equation. Why? Because the concentration of a pure solid is constant and is already baked into the $K_{sp}$ value itself!

### Finding the Molar Solubility
Let's define the molar solubility of silver iodate as $S$ (in $	ext{mol/L}$). When $S$ moles of $	ext{AgIO}_3$ dissolve, they produce $S$ moles of $	ext{Ag}^+$ ions and $S$ moles of $	ext{IO}_3^-$ ions.

Substituting these into our $K_{sp}$ expression gives:

$$K_{sp} = S 	imes S = S^2$$

We are given that $K_{sp} = 1.0 	imes 10^{-8}$. Let's plug that in and solve for $S$:

$$1.0 	imes 10^{-8} = S^2$$

Taking the square root of both sides, we find:

$$S = \sqrt{1.0 	imes 10^{-8}} = 1.0 	imes 10^{-4} 	ext{ mol/L}$$

This tells us that in one full liter ($1000	ext{ mL}$) of a saturated solution, exactly $1.0 	imes 10^{-4}$ moles of $	ext{AgIO}_3$ are dissolved.

### Final Calculation: Scaling Down to 100 mL
The question throws a slight curveball: it asks for the mass in just $100	ext{ mL}$ of the solution, not a full liter.

Since $100	ext{ mL}$ is exactly one-tenth of a liter ($1000	ext{ mL}$), the number of moles dissolved in $100	ext{ mL}$ will also be one-tenth of the molar solubility:

$$	ext{Moles in 100 mL} = \frac{1.0 	imes 10^{-4}}{10} = 1.0 	imes 10^{-5} 	ext{ mol}$$

Finally, we need to convert these moles into grams. We do this by multiplying the number of moles by the molar mass of $	ext{AgIO}_3$, which is given as $283	ext{ g/mol}$:

$$	ext{Mass} = 	ext{Moles} 	imes 	ext{Molar Mass}$$

$$	ext{Mass} = 1.0 	imes 10^{-5} 	ext{ mol} 	imes 283 	ext{ g/mol}$$

$$	ext{Mass} = 2.83 	imes 10^{-3} 	ext{ g}$$

And there we have it! The mass of silver iodate dissolved in $100	ext{ mL}$ of its saturated solution is **$2.83 	imes 10^{-3}	ext{ g}$**, which perfectly matches option (b).

Select Answer:

Visualized Solution

$Dissolution Equilibrium$

$Solubility Product (K_{s p})$

$Substituting Values$

$Calculating Solubility (S)$

$Moles in 100 mL$

$Mass of AgIO_{3}$

$Common Ion Effect$

The Sigma Insight: Solubility Product and Common Ion Effect

Analyzing the Setup Imagine a beaker filled with water

The Master Equation

Finding the Molar Solubility Let's define the molar solubility of silver iodate as $S$ (in $mol/L$ )

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solubility Product (Ksp​)

.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; } Calculating Solubility (S)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Moles in 100 mL

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Mass of AgIO3​

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