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Question: pH of saturated solution of \(Ca{{(OH)}_{2}}\) is 9. The solubility product (\({{K}_{sp}}\) ) of \(C...

pH of saturated solution of Ca(OH)2Ca{{(OH)}_{2}} is 9. The solubility product (Ksp{{K}_{sp}} ) of Ca(OH)2Ca{{(OH)}_{2}}is:
(A) 0.5×10−150.5\times {{10}^{-15}}
(B) 0.25×10−100.25\times {{10}^{-10}}
(C) 0.125×10−150.125\times {{10}^{-15}}
(D) 0.5×10100.5\times {{10}^{10}}

Explanation

Solution

The solubility product constant is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble substance means it has a higher solubility product value. Solubility products were only applicable for sparingly soluble ionic compounds. The state of the solution when no solution is added after this state resulting precipitating solution, such a mixture is called a saturated solution.

Complete step by step solution:
Adding solute a solution until where the solute appears as solid crystals formed a highly saturated solution.
For example, adding a tablespoon of sugar to a container of sugar, initially, sugar dissolves with continuous stirring and more sugar has added a point where no amount of sugar is required for stirring to dissolve. At last, added sugar remains as solid on the bottom of the container, then the solution is said to be saturated.
Given saturated solution is Ca(OH)2Ca{{(OH)}_{2}}
The pH of the given saturated solution calcium hydroxide = 9
pH+pOH=14pH + pOH = 14
Then, pOH=14–pH=14–9=5pOH = 14 – pH = 14 – 9 = 5
pOH=−log⁡[OH−]pOH=-\log [O{{H}^{-}}]
⇒5=−log⁡[OH−]\Rightarrow 5=-\log [O{{H}^{-}}]
[OH−]=10−5M[O{{H}^{-}}]={{10}^{-5}}M
Ca(OH)2(aq)→Ca+2(aq)+2OH−(aq)Ca{{(OH)}_{2}}(aq)\to C{{a}^{+2}}(aq)+2O{{H}^{-}}(aq) -- (1)
From the above reaction, 1 mole of calcium hydroxide dissociates into one mole of calcium ion and 2 moles of hydroxide ions.
Then, [OH−]=10−5M[O{{H}^{-}}]={{10}^{-5}}M
[Ca+2]=[OH−]2=0.5×10−5M[C{{a}^{+2}}]=\dfrac{[O{{H}^{-}}]}{2}=0.5\times {{10}^{-5}}M
Solubility product of given equation,
Ksp=[Ca+2][OH−]2{{K}_{sp}}=[C{{a}^{+2}}]{{[O{{H}^{-}}]}^{2}}
Ksp=0.5×10−5×(10−5)2=0.5×10−15{{K}_{sp}}=0.5\times {{10}^{-5}}\times {{({{10}^{-5}})}^{2}}=0.5\times {{10}^{-15}}
Hence, the solubility product (Ksp{{K}_{sp}} ) Ca(OH)2Ca{{(OH)}_{2}} is 0.5×10−150.5\times {{10}^{-15}}

The correct answer is option A.

Note: Solubility products cannot be used for normally soluble compounds like sodium chloride, sodium nitrate, etc. interactions between the ions in the solution interfere with the simple equilibrium. The solubility product is a value that you get when the solution is saturated. If there is any solid present, cannot dissolve any more solid than there is a saturated solution.