The solubility of KCl is 3.7 M at 20 °C. Two beakers each contain 100. mL of saturated KCl solution: 100. mL of 4.0 M HCl is add

Answer:

(a) 13.69

(b) i beaker 1: 0g

    ii beaker 2: 0g

Explanation:

a. The KCl solubility equilibrium equation is

KCl(s)  ⇄  K⁺(aq)  +  Cl⁻(aq)

A 3.7M KCl solution contains equal amounts of K ions and Cl ions hence

the ion-product formula can be outlined as

Ksp = [K⁺][Cl⁻]

       = [3.7][3.7]

       = 13.69

b. From the two beakers, each holding 100 mL and 3.7M KCl

the molarity of K⁺ = molarity of Cl⁻ = moles of KCl = 3.7moles in 1L,

noting that 3.7M represents 3.7 mol in 1L or 1000mL or 1000 cm³.

Calculating the total moles contained in 100 mL:

3.7moles/Liter * 100 mL

= 0.37moles K⁺ = 0.37moles Cl⁻.

A 4.0 M HCl solution contains

in 100mL.

An 8.0M HCl solution contains

in 100mL.

In the first beaker, 100 mL of 4M HCl is combined with 100 mL of 3.7M KCl:

Total moles of Cl⁻ (0.4 + 0.37) equals 0.77 moles.

While total moles of K⁺ remain 0.37 moles.

The total solution volume = (100mL + 100mL) = 200mL/1000mL = 0.2L.

Thus moles of Cl⁻ calculated per liter = 0.77moles/0.2L = 3.85M Cl⁻.

Moles of K⁺ calculated per liter = 0.37moles/0.2L = 1.85M K⁺.

For precipitation to take place, Qsp should be equal to or exceed Ksp, which means:

Qsp ≥ Ksp

Qsp = [K][Cl] = [1.85][3.85] = 7.12, which remains below 13.69 (Ksp).

Consequently, no KCl will precipitate in the first beaker, and since there’s no precipitate, calculating the mass that would have formed is unnecessary.

Thus, the result is 0g.

(bii) In the second beaker, 100 mL of 8M HCl is combined with 100 mL of 3.7M KCl:

Total moles of Cl⁻ (0.8 + 0.37) total to 1.17 moles.

With total moles of K⁺ remaining at 0.37 moles.

Providing the total solution volume = (100mL + 100mL) = 200mL/1000mL = 0.2L.

Total moles of Cl⁻ per liter = 1.17moles/0.2L = 5.85M Cl⁻.

Total moles of K⁺ per liter = 0.37moles/0.2L = 1.85M K⁺.

For precipitation to happen, Qsp needs to be at minimum equal to Ksp, which implies:

Qsp ≥ Ksp

Qsp = [K][Cl] = [1.85][5.85] = 10.82 which, again, is below 13.69 (Ksp).

Thus, no KCl will precipitate in the second beaker; thus, calculating mass for precipitation is also rendered unnecessary.

The answer here is also 0g.

Answer:

a) The ion-product constant Ksp was determined to be 13.69

Explanation:

Terminology

The Qsp for an ionic solid dissolving reflects the solubility product of ions in solution.

Ksp, in contrast, defines the equilibrium state solubility product of these ions in solution when in balance with the dissolving solid.

It’s important to note that if Qsp exceeds Ksp at a particular temperature, precipitation will occur, causing the equilibrium to shift left to maintain balance (Ksp).

Steps to Solve:

To calculate this:

1. Replace the molar solubility of KCl into the ion-product equation to identify the Ksp of KCl.

2. Assess the total concentration of potassium chloride ions in each beaker after HCl has been added, taking into consideration initial moles and those added.

3. Calculate Qsp to determine whether it surpasses Ksp. If Qsp is below Ksp, no precipitation occurs.

Thus, the equilibrium equation for KCl can be expressed as:

KCL_(s) ---> K+(aq) + Cl- (aq)

The provided KCl solubility is 3.7 M.

Ksp= [K+][Cl-] = (3.7)(3.7) =13.69

Therefore, the Ksp was found to be equal to 13.69.

In pure water, KCl

Ksp =13.69 KCl =[K+][Cl-]

Let x represent the molar solubility [K+]/[Cl-]:. x, x

Ksp =13.69 = [K+][Cl-] = (x)(x) = x²

x= √ 13.69 = 3.7 M moles of KCl necessary for a 100mL saturated solution

37M moles/L

The Ksp was determined to be equal to 13.69.

4.0 M HCl = KCl =[K+][Cl-]

Let y signify the molar solubility:. y, y+4

Ksp =13.69= [K+][Cl-] = (y)(y*+4)

* - as a general guideline

Ksp =13.69= [K+][Cl-] = (y)(y*+4)= y(4)

13.69=4y:. y= 3.42 moles/100mL

y= 34.2moles/L

8 M HCl = KCl =[K+][Cl-]

Let b denote the molar solubility:. B, b+8

Ksp =13.69= [K+][Cl-] = (b)(b*+8)

* - as a general guideline

Ksp =13.69= [K+][Cl-] = (b)(b*+8)= b(8)

13.69=8b:. b= 1.71 moles/100mL

17.1 moles/L

Thus, in a solution containing a common ion, the compound's solubility significantly decreases.