Q) The Ksp of Ca(OH)2 is 4.42*(1/100000) at 25degree centigrade. 500ml of saturated solution of calcium hydroxide is mixed with equal volume of 0.4M NaOH. How much Ca(OH)2 is presepited.

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To determine how much calcium hydroxide, Ca(OH)2, will precipitate when a saturated solution of Ca(OH)2 is mixed with an equal volume of 0.4 M sodium hydroxide (NaOH), we need to analyze the solubility product constant (Ksp) and the changes in concentration that occur during the mixing process.

Understanding the Components

First, let's break down the information provided:

• The Ksp of Ca(OH)2 is given as 4.42 × 10^-6 at 25 degrees Celsius.
• You have 500 mL of a saturated solution of Ca(OH)2.
• This saturated solution is mixed with 500 mL of 0.4 M NaOH.

Calculating the Saturation Concentration of Ca(OH)2

The Ksp expression for calcium hydroxide is:

Ksp = [Ca^2+][OH^-]^2

Let’s denote the solubility of Ca(OH)2 in mol/L as 's'. In a saturated solution, the concentration of calcium ions [Ca^2+] will be 's', and the concentration of hydroxide ions [OH^-] will be '2s' (since each formula unit of Ca(OH)2 produces two hydroxide ions).

Substituting these into the Ksp expression gives:

Ksp = s(2s)^2 = 4s^3

Setting this equal to the Ksp value:

4s^3 = 4.42 × 10^-6

Solving for 's':

s^3 = 1.105 × 10^-6

s = (1.105 × 10^-6)^(1/3) ≈ 0.0102 M

This means the concentration of Ca(OH)2 in the saturated solution is approximately 0.0102 M.

Mixing the Solutions

When you mix 500 mL of saturated Ca(OH)2 with 500 mL of 0.4 M NaOH, the total volume becomes 1 L. The concentration of NaOH in the mixed solution can be calculated as follows:

[NaOH] = (0.4 M × 0.5 L) / 1 L = 0.2 M

In the mixed solution, the concentration of hydroxide ions from NaOH will be 0.2 M, and we also have the contribution from the saturated Ca(OH)2 solution.

Calculating the Total Hydroxide Ion Concentration

The total concentration of hydroxide ions [OH^-] in the mixed solution will be:

[OH^-] = 2s + [NaOH] = 2(0.0102) + 0.2 = 0.0204 + 0.2 = 0.2204 M

Determining Precipitation of Ca(OH)2

Now, we need to find out if the new concentration of hydroxide ions exceeds the solubility limit of Ca(OH)2. We can use the Ksp expression again:

Ksp = [Ca^2+][OH^-]^2

Let’s denote the concentration of calcium ions as 'x'. The Ksp expression becomes:

4.42 × 10^-6 = x(0.2204)^2

Calculating [OH^-]^2:

(0.2204)^2 ≈ 0.0482

Now substituting this back into the Ksp equation:

4.42 × 10^-6 = x(0.0482)

Solving for 'x':

x = (4.42 × 10^-6) / (0.0482) ≈ 9.16 × 10^-5 M

Finding the Amount of Precipitated Ca(OH)2

Since the initial concentration of calcium ions in the saturated solution was 0.0102 M, and the new concentration after mixing is approximately 9.16 × 10^-5 M, we can find the amount of Ca(OH)2 that precipitates:

The difference in concentration of Ca^2+ ions before and after mixing gives us:

Δ[Ca^2+] = 0.0102 M - 9.16 × 10^-5 M ≈ 0.0101 M

Now, to find the amount of Ca(OH)2 that precipitates, we can use the molarity and the volume of the solution:

Volume of solution = 1 L

Amount of Ca(OH)2 precipitated = Δ[Ca^2+] × Volume = 0.0101 mol/L × 1 L = 0.0101 mol

Final Calculation

To convert moles of Ca(OH)2 to grams, we use the molar mass of Ca(OH)2, which is approximately 74.09 g/mol:

Mass = 0.0101 mol × 74.09 g/mol ≈ 0.747 g

In summary, when the saturated solution of calcium hydroxide is mixed with 0.4 M sodium hydroxide, approximately 0.747 grams of Ca(OH)2 will precipitate out of the solution.