Answer:
The K sp Value is [tex]K_{sp}=7.40[/tex]
Explanation:
From the question we are told that
The of [tex]KClO_3[/tex] is = 122.5 g/ mol
The mass of [tex]KClO_3[/tex] dissolved is [tex]m_s = 4.0g[/tex]
The volume of solution is [tex]V_s = 12mL = 12*10^{-3}L[/tex]
The number of moles of [tex]KClO_3[/tex] is mathematically evaluated as
[tex]No \ of \ moles \ = \frac{mass }{Molar \ mass}[/tex]
Substituting values
[tex]= \frac{4}{122.5}[/tex]
[tex]=0.0327\ moles[/tex]
Generally concentration is mathematically represented as
[tex]concentration = \frac{No \ of \ moles}{volume }[/tex]
For [tex]KClO_3[/tex]
[tex]Z= \frac{0.0327}{12*10^{-3}}[/tex]
[tex]=2.72 \ mol/L[/tex]
The dissociation reaction of [tex]KClO_3[/tex] is
[tex]KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}[/tex]
The solubility product constant is mathematically represented as
[tex]K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }[/tex]
Since there is no ionic reactant we have
[tex]K_{sp} = [k^+] [ClO_3^-][/tex]
[tex]= Z^2[/tex]
[tex]= 2.72^2[/tex]
[tex]K_{sp}=7.40[/tex]
A solution is prepared by dissolving 4.00 g of KClO₃ in 12 mL of water at 85 °C. When the solution cools to 74 °C, the solid begins to appear. At that temperature, the Ksp for KClO₃ is 7.3.
At 74 °C, up to 4.00 g of KClO₃ can be dissolved in 12 mL of water. We can calculate the molar solubility of KClO₃ (S) using the following expression.
[tex]S = \frac{mKClO_3}{M(KClO_3)\times liters\ solution } = \frac{4.00g}{122.5g/mol\times 0.012L } = 2.7 M[/tex]
We can relate the molar solubility (S) with the solubility product constant (Ksp) through the ICE chart for the solution of KClO₃.
KClO₃(s) ⇄ K⁺(aq) + ClO₃⁻(aq)
I 0 0
C +S +S
E S S
The solubility product constant (Ksp) is:
[tex]Ksp=[K^{+} ][ClO_3^{-} ]= S.S=S^{2} = 2.7^{2} = 7.3[/tex]
A solution is prepared by dissolving 4.00 g of KClO₃ in 12 mL of water at 85 °C. When the solution cools to 74 °C, the solid begins to appear. At that temperature, the Ksp for KClO₃ is 7.3.
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