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

The Ksp of KClO₃ at 74°C is calculated using the saturated solution's molar solubility. By dissolving 4.00 g of KClO₃ in water and calculating the molarity of this solution, the Ksp is found to be 7.4028 M².

Step-by-step explanation:

To find the Ksp for KClO₃ at 74°C, we first need to determine the molar solubility of the compound at this temperature. Since 4.00 g of KClO₃ begins to form a solid at 74°C in 12 mL of water, this is the solubility limit at that temperature. With the molar mass of KClO₃ being 122.5 g/mol, we can convert the mass of KClO₃ to moles:

Moles of KClO₃ = 4.00 g / 122.5 g/mol = 0.03265 mol

Next, we calculate the molarity of the saturated solution by dividing the moles by the volume of water in liters (12 mL = 0.012 L):

Molarity of KClO₃ = 0.03265 mol / 0.012 L = 2.7208 M

This molarity is the molar solubility of KClO₃ at 74°C. The dissolution of KClO₃ can be represented as KClO₃(s) ↔ K+(aq) + ClO₃⁻(aq). Thus, for each mole of KClO₃ that dissolves, 1 mole of K+ and 1 mole of ClO₃⁻ are produced.

The Ksp expression is: Ksp = [K+] [ClO3-]. Since the concentrations are equal, the Ksp is simply the square of the molar solubility:

Ksp of KClO₃ at 74°C = (2.7208 M)² = 7.4028 M²

Answer 2

The K sp Value is
`K_(sp)=7.40`

Step-by-step 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 \ = (mass )/(Molar \ mass)`

Substituting values

`= (4)/(122.5)`

`=0.0327\ moles`

Generally concentration is mathematically represented as

`concentration = (No \ of \ moles)/(volume )`

For
`KClO_3`

`Z= (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) = (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`