Calculate the molar solubility of calcium hydroxide in a solution buffered at each of the following ph's: ? ph 5 ph 7 ph 8

The solubility constant at 25°C is [1]
Ksp = 5.02×10^–6 Let s be the molar solubility of Ca(OH)₂.[OH⁻] = 10^(pH - 14)

hence
Ksp = s ∙ (10^(pH - 14))² = s ∙ 10^(2∙pH - 28)
=>
s = Ksp / 10^(2∙pH - 28) = Ksp ∙ 10^(28 - 2∙pH)

at pH=5
s = 5.02×10^–6 ∙ 10^(28 - 2∙5) = 5.02×10^–6 ∙ 10^(18) = 5.02×10^12 mol/L

at pH=7
s = 5.02×10^–6 ∙ 10^(28 - 2∙7) = 5.02×10^–6 ∙ 10^(14) = 5.02×10^8 mol/L

at pH=8
s = 5.02×10^–6 ∙ 10^(28 - 2∙7) = 5.02×10^–6 ∙ 10^(12) = 5.02×10^6 mol/L

The number of moles of a given substance dissolved in solution in liter's before its saturation point is called molar solubility.

Dissociation reaction of calcium hydroxide is shown as:

The solubility product of is mathematically expressed as:

The solubility constant is given by:

Assuming, molar solubility of

Hence,

Solving further:

At pH 5:

At pH 7:

At pH 8:

Therefore molar solubility's are:

pH 5 ,

pH 7  

pH 8

To learn more about molar solubility follow the link:

link

Answer trigonal pyramidal;

The answer is conduction