Question

Suppose you are studying the K sp of K C l O 3 , which has a molar mass of 122.5 g/mol, at multiple temperatures. You dissolve 4.00 g of K C l O 3 in 12 mL of water at 85 oC and cool the solution. At 74 oC, a solid begins to appear. What is the K sp of K C l O 3 at 74 oC?

Answer 1

Answer:

Ksp of KClO3 is 7.40

Explanation:

Step 1: Data given

Mass of KClO3 = 4.00 grams

Molar mass of KClO3 = 122.5 g/mol

Volume of water = 12 mL = 0.012 L

Temperature = 85 °C

Step 2: The balanced equation

KClO3 →K+ + ClO3-

Step 3: Calculate moles of KClO3

Moles KClO3 = mass KClO3 / molar mass KClO3

Moles KClO3 = 4.00 grams / 122.5 g/mol

Moles KClO3 = 0.03265 moles

Step 4: Calculate concentration

Concentration = moles / volume

[KClO3] = 0.03265 moles / 0.012 L

[KClO3] = 2.72 M

For 1 mKsp of KClO3 is 7.40ol KClO3 we have 1 mol K+ and 1 mol ClO3-

[KClO3] = [K+] = [ClO3-] = 2.72 M

Step 5: Calculate Ksp

Ksp = [K+][ClO3-]

Ksp = 2.72 * 2.72

Ksp =  7.40

Answer 2

Answer:

The K sp Value is  $K_{sp}=7.40$

Explanation:

From the question we are told that

   The of $KClO_3$ is = 122.5 g/ mol

    The mass of $KClO_3$ dissolved is $m_s = 4.0g$

    The volume of solution is  $V_s = 12mL = 12*10^{-3}L$

The number of moles of $KClO_3$ is mathematically evaluated as

           $No \ of \ moles \ = \frac{mass }{Molar \ mass}$

Substituting values

                                  $= \frac{4}{122.5}$

                                  $=0.0327\ moles$

Generally concentration is mathematically represented as

         $concentration = \frac{No \ of \ moles}{volume }$

For $KClO_3$        

               $Z= \frac{0.0327}{12*10^{-3}}$

                              $=2.72 \ mol/L$

The dissociation reaction of $KClO_3$  is

         $KClO_3 \ ----> K^{+}_{(aq)} + ClO_3^-_{(aq)}$

The solubility product constant is mathematically represented as

                   $K_{sp} = \frac{concentration of ionic product }{concentration of ionic reactant }$

Since there is no ionic reactant we have

                  $K_{sp} = [k^+] [ClO_3^-]$

                          $= Z^2$

                          $= 2.72^2$

                          $K_{sp}=7.40$