<span>The solubility constant at 25°C is [1]
Ksp = 5.02×10^–6 </span>Let s be the molar solubility of Ca(OH)₂.<span>[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 </span>
Blue has a shorter wave length
Answer:
This is not a redox reaction. None of the species are reduced and none are oxidized
Explanation:
In a redox reaction at least one species must be oxidized and another reduced. You see this by a change in oxidation number. In this question the oxidation numbers are the same before and after the reaction.
Ca 2
H -1
Na +1
I -1
Have about 5 beakers all with different temperatures of water. Put in a teaspoon of salt at a time and when it stops dissolving stop adding and record how much salt it took. It should be more salt as the temperature rises. The independent variable is the waters temperature. The dependent variable is how much salt is used. Make sure that there is the same amount of water in each beaker. Or else it won’t work.