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.

To determine how much calcium hydroxide, Ca(OH)₂, will precipitate when a saturated solution is mixed with sodium hydroxide (NaOH), we first need to understand the solubility product constant (Ksp) and how it influences the solubility of the compound in the solution.

Understanding Ksp and Saturation

The Ksp of Ca(OH)₂ is given as 4.42 × 10⁻⁵ at 25 degrees Celsius. This value indicates the extent to which calcium hydroxide can dissolve in water. The dissolution of Ca(OH)₂ can be represented by the following equilibrium reaction:

Ca(OH)₂ (s) ⇌ Ca²⁺ (aq) + 2 OH⁻ (aq)

The Ksp expression for this equilibrium is:

Ksp = [Ca²⁺][OH⁻]²

Calculating Ion Concentrations in Saturated Solution

In a saturated solution of Ca(OH)₂, let’s denote the solubility of Ca(OH)₂ as 's'. Therefore, at saturation:

• [Ca²⁺] = s
• [OH⁻] = 2s

Substituting these into the Ksp expression gives:

Ksp = s(2s)² = 4s³

Setting this equal to the Ksp value:

4s³ = 4.42 × 10⁻⁵

From this, we can solve for 's':

s³ = 1.105 × 10⁻⁵

s ≈ 0.0222 M

Mixing with NaOH

Now, we have a saturated solution of Ca(OH)₂ with a concentration of approximately 0.0222 M. When we mix this with an equal volume of 0.4 M NaOH, we need to consider the total concentration of hydroxide ions (OH⁻) in the resulting solution.

In the 500 mL of saturated Ca(OH)₂ solution, the concentration of OH⁻ ions is:

[OH⁻] from Ca(OH)₂ = 2 × 0.0222 M = 0.0444 M

In the 500 mL of 0.4 M NaOH solution, the concentration of OH⁻ ions is:

[OH⁻] from NaOH = 0.4 M

When mixed, the total volume becomes 1 L (1000 mL), and the total concentration of OH⁻ ions in the mixture is:

[OH⁻] total = (0.0444 M × 0.5 L + 0.4 M × 0.5 L) / 1 L = (0.0222 + 0.2) M = 0.2222 M

Determining Precipitation of Ca(OH)₂

Now, we need to check if this concentration of OH⁻ exceeds the solubility limit of Ca(OH)₂. The solubility of Ca(OH)₂ is determined by its Ksp:

From the Ksp expression, we can find the maximum concentration of OH⁻ that can exist without precipitating Ca(OH)₂:

Ksp = [Ca²⁺][OH⁻]²

Assuming 's' is the concentration of Ca²⁺ at saturation (0.0222 M), we can rearrange the Ksp expression:

4.42 × 10⁻⁵ = 0.0222 × [OH⁻]²

Solving for [OH⁻]:

[OH⁻]² = (4.42 × 10⁻⁵) / 0.0222

[OH⁻]² ≈ 1.989 × 10⁻³

[OH⁻] ≈ 0.0446 M

Since the total concentration of OH⁻ in the mixed solution (0.2222 M) is much greater than 0.0446 M, Ca(OH)₂ will precipitate.

Calculating the Amount of Precipitate

To find out how much Ca(OH)₂ precipitates, we can determine how much Ca²⁺ can remain in solution at the new concentration of OH⁻:

Using the Ksp expression again:

4.42 × 10⁻⁵ = [Ca²⁺](0.2222)²

Solving for [Ca²⁺]:

[Ca²⁺] = 4.42 × 10⁻⁵ / (0.2222)²

[Ca²⁺] ≈ 0.00089 M

Now, we can find the amount of Ca²⁺ that precipitates:

Initially, the concentration of Ca²⁺ was 0.0222 M in the saturated solution. After mixing, it drops to 0.00089 M. The difference gives us the concentration of Ca²⁺ that has precipitated:

Precipitated Ca²⁺ = 0.0222 M - 0.00089 M ≈ 0.02131 M

Since we have 1 L of the mixed solution, the moles of Ca(OH)₂ that precipitate is:

0.02131 moles/L × 1 L = 0.02131 moles

Final Calculation of Mass

To find the mass of the precipitated Ca(OH)₂, we use its molar mass (approximately 74.09 g/mol):

Mass = moles × molar mass

Mass = 0.02131 moles × 74.09 g/mol ≈ 1.577 g

Thus, approximately 1.577 grams of Ca(OH)₂ will precipitate when the saturated solution is mixed with the NaOH solution.