Respuesta :
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
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:
[tex]\begin{aligned}\rm Ca(OH)_{2} (s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{-}(aq) \end{aligned}[/tex]
The solubility product of [tex]\text{Ca(OH)}_{2}[/tex] is mathematically expressed as:
[tex]\text{Ksp} &= \rm [Ca^{2+}] [OH^{-}]^{2}[/tex]
The solubility constant is given by:
[tex]\text {Ksp} = 5.02 \times 10^{-6}[/tex]
Assuming, molar solubility of [tex]\rm Ca(OH)_{2}.[OH^{-}] = 10^{(pH - 14)}[/tex]
Hence,
[tex]\begin{aligned} \rm Ksp &= \rm s (10^{(pH - 14)})^{2} \\\\&= \rm s 10^{(2(pH) - 28)} \end{aligned}[/tex]
Solving further:
[tex]\begin{aligned} \rm s &= \rm \dfrac {Ksp }{10^{(2(pH) - 28)} }\\\\&= \rm Ksp \times 10^{(28 - 2(pH))}\end{aligned}[/tex]
- At pH 5:
[tex]\begin{aligned} \rm s & = 5.02 \times 10^{-6 } \times 10^{(28 - 2(5))}\\&= 5.02 \times 10^{-6 } \times 10^{(18)} \\&= 5.02 \times 10^{12} \rm mol/L \end{aligned}[/tex]
- At pH 7:
[tex]\begin{aligned} \rm s & = 5.02 \times 10^{-6 } \times 10^{(28 - 2(7))}\\&= 5.02 \times 10^{-6 } \times 10^{(14)} \\&= 5.02 \times 10^{8} \rm mol/L \end{aligned}[/tex]
- At pH 8:
[tex]\begin{aligned} \rm s & = 5.02 \times 10^{-6 } \times 10^{(28 - 2(8))}\\&= 5.02 \times 10^{-6 } \times 10^{(12)} \\&= 5.02 \times 10^{6} \rm mol/L \end{aligned}[/tex]
Therefore molar solubility's are:
pH 5 [tex]\begin{aligned} &= 5.02 \times 10^{12} \rm mol/L \end{aligned}[/tex],
pH 7 [tex]\begin{aligned} &= 5.02 \times 10^{8} \rm mol/L \end{aligned}[/tex]
pH 8 [tex]\begin{aligned} &= 5.02 \times 10^{6} \rm mol/L \end{aligned}[/tex]
To learn more about molar solubility follow the link:
https://brainly.com/question/1525823
