A mixture of 100 mmol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume wa

Q: Why do we ignore the remaining calcium hydroxide when calculating the final OH- concentration?

We ignore it because of the Common Ion Effect. The strong base NaOH produced in the reaction provides a high concentration of OH- ions (0.28 M). This high concentration suppresses the dissociation of the sparingly soluble Ca(OH)2 (which has a very low Ksp), meaning it contributes a negligible amount of additional OH- ions.

Q: How did we get 14 millimoles of sodium sulphate?

We divided the given mass of sodium sulphate (2 g) by its molar mass (143 g/mol). This gives approximately 0.014 moles, which is equal to 14 millimoles.

Q: Why is the volume taken as 100 mL for the final concentration?

The problem explicitly states that after the mixture was dissolved in water, 'the volume was made upto 100 mL'. Therefore, the final volume of the entire solution is 100 mL.

Q: What does the Ksp value tell us here?

The Ksp (Solubility Product Constant) value indicates how much of a sparingly soluble salt can dissolve. A very small Ksp like 5.5 x 10^-6 confirms that Ca(OH)2 does not dissolve easily, especially in the presence of a common ion.

Q: Can I use molarity instead of millimoles for the calculations?

Yes, you absolutely can. However, using millimoles is often easier and less prone to decimal errors when the volume is given in milliliters, as Molarity = millimoles / milliliters.

Question

A mixture of 100 mmol of $	ext{Ca(OH)}_2$ and 2 g of sodium sulphate was dissolved in water and the volume was made upto 100 mL. The mass of calcium sulphate formed and the concentration of $	ext{OH}^-$ in resulting solution, respectively, are : (Molar mass of $	ext{Ca(OH)}_2$, $	ext{Na}_2	ext{SO}_4$ and $	ext{CaSO}_4$ are 74, 143 and $136	ext{ g mol}^{-1}$, respectively; $K_{sp}$ of $	ext{Ca(OH)}_2$ is $5.5 	imes 10^{-6}$)

Options:

(A) $13.6	ext{ g}, 0.28	ext{ mol L}^{-1}$

(B) $1.9	ext{ g}, 0.28	ext{ mol L}^{-1}$

(C) $13.6	ext{ g}, 0.14	ext{ mol L}^{-1}$

(D) $1.9	ext{ g}, 0.14	ext{ mol L}^{-1}$

Sigma Percentile · Accepted Answer

### The Case of the Limiting Reagent and the Common Ion Effect

Imagine you are a chemical detective, and you've just been handed a beaker containing a mysterious mixture. Inside this beaker, $100 	ext{ mmol}$ of calcium hydroxide, $	ext{Ca(OH)}_2$, and $2 	ext{ g}$ of sodium sulphate, $	ext{Na}_2	ext{SO}_4$, are swirling together in $100 	ext{ mL}$ of water. Your mission? To determine exactly how much solid precipitate forms and the final concentration of the hydroxide ions floating in the solution.

Let's break this down step-by-step, just like a true chemist.

### Setting the Stage: The Reactants

First, we need to understand the chemical conversation happening in our beaker. When calcium hydroxide meets sodium sulphate, they undergo a classic double displacement reaction. The calcium pairs up with the sulphate to form a solid precipitate, while the sodium pairs with the hydroxide to remain dissolved in the aqueous solution.

The balanced chemical equation is:

$$	ext{Ca(OH)}_2 + 	ext{Na}_2	ext{SO}_4 ightarrow 	ext{CaSO}_4 \downarrow + 2	ext{NaOH}$$

Notice the stoichiometry: one mole of $	ext{Ca(OH)}_2$ reacts with exactly one mole of $	ext{Na}_2	ext{SO}_4$ to produce one mole of $	ext{CaSO}_4$ and two moles of $	ext{NaOH}$.

### The Limiting Reagent: Who Runs Out First?

To predict the outcome, we must find out which reactant will be completely consumed first. This is our **limiting reagent**. We already know we have $100 	ext{ mmol}$ of $	ext{Ca(OH)}_2$. But what about $	ext{Na}_2	ext{SO}_4$?

We are given its mass as $2 	ext{ g}$. To convert this to moles, we divide by its molar mass ($143 	ext{ g/mol}$):

$$n_{	ext{Na}_2	ext{SO}_4} = \frac{2 	ext{ g}}{143 	ext{ g/mol}} \approx 0.014 	ext{ mol} = 14 	ext{ mmol}$$

Comparing the two, we have $100 	ext{ mmol}$ of $	ext{Ca(OH)}_2$ but only $14 	ext{ mmol}$ of $	ext{Na}_2	ext{SO}_4$. Since they react in a 1:1 ratio, the $	ext{Na}_2	ext{SO}_4$ will run out long before the $	ext{Ca(OH)}_2$ does. Therefore, **$	ext{Na}_2	ext{SO}_4$ is our limiting reagent** and will dictate the amount of products formed.

### Calculating the Yield: Mass of the Precipitate

Because $14 	ext{ mmol}$ of $	ext{Na}_2	ext{SO}_4$ reacts completely, it will produce exactly $14 	ext{ mmol}$ of our precipitate, calcium sulphate ($	ext{CaSO}_4$).

To find the mass of this precipitate, we multiply the moles by its molar mass ($136 	ext{ g/mol}$):

$$W_{	ext{CaSO}_4} = 14 	imes 10^{-3} 	ext{ mol} 	imes 136 	ext{ g/mol} = 1.904 	ext{ g}$$

Rounding to one decimal place, we get **$1.9 	ext{ g}$** of $	ext{CaSO}_4$.

### The Hydroxide Concentration: A Strong Base Emerges

Now, let's turn our attention to the hydroxide ions. According to our balanced equation, every mole of $	ext{Na}_2	ext{SO}_4$ that reacts produces *two* moles of $	ext{NaOH}$.

So, $14 	ext{ mmol}$ of $	ext{Na}_2	ext{SO}_4$ will yield:

$$n_{	ext{NaOH}} = 2 	imes 14 	ext{ mmol} = 28 	ext{ mmol}$$

Since $	ext{NaOH}$ is a strong base, it dissociates completely, giving us $28 	ext{ mmol}$ of $	ext{OH}^-$ ions. To find the concentration (Molarity), we divide these millimoles by the total volume of the solution ($100 	ext{ mL}$):

$$[	ext{OH}^-] = \frac{28 	ext{ mmol}}{100 	ext{ mL}} = 0.28 	ext{ M}$$

### The Plot Twist: Why the $K_{sp}$?

You might be thinking, "Wait a minute! We started with $100 	ext{ mmol}$ of $	ext{Ca(OH)}_2$ and only used $14 	ext{ mmol}$. What about the remaining $86 	ext{ mmol}$? Doesn't it also release $	ext{OH}^-$ ions?"

This is where the problem throws a brilliant curveball by providing the solubility product constant, $K_{sp}$, of $	ext{Ca(OH)}_2$ ($5.5 	imes 10^{-6}$). This tiny value tells us that $	ext{Ca(OH)}_2$ is only sparingly soluble.

Furthermore, we already have a massive concentration of $	ext{OH}^-$ ($0.28 	ext{ M}$) floating in the solution from the strong base $	ext{NaOH}$. According to Le Chatelier's Principle, this high concentration of a common ion will push the dissociation equilibrium of $	ext{Ca(OH)}_2$ far to the left. This phenomenon is known as the **Common Ion Effect**.

Because of this effect, the remaining $86 	ext{ mmol}$ of $	ext{Ca(OH)}_2$ will remain almost entirely undissolved as a solid. Its contribution to the total $	ext{OH}^-$ concentration is so infinitesimally small that we can safely ignore it.

Thus, our final answers stand strong: **$1.9 	ext{ g}$** of precipitate and a **$0.28 	ext{ M}$** concentration of hydroxide ions.

Select Answer:

Visualized Solution

\text{Chemical Reaction}

\text{Initial Millimoles}

\text{Limiting Reagent}

\text{Mass of CaSO}_4 \text{ Formed}

\text{Concentration of OH}^-

\text{Role of } K_{sp}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Case of the Limiting Reagent and the Common Ion Effect

Setting the Stage

The Limiting Reagent

Calculating the Yield

The Hydroxide Concentration

The Plot Twist

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