Respuesta :
Answer:
D) Ca(OH)₂ will not precipitate because Q < Ksp
Explanation:
Here we have first a chemical reaction in which Ca(OH)₂ is produced:
CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂
Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.
The solubility product constant for Ca(OH)₂ is:
Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)
Ksp = [Ca²⁺][OH⁻]²
and the reaction quotient Q:
Q = [Ca²⁺][OH⁻]²
So by comparing Q with Ksp we will be able to determine if a precipitate will form.
From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.
mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂
the concentration of ions will be:
[Ca²⁺ ] = 0.001 mol / L 0.001 M
[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)
Now we can calculate the reaction quotient.
Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹
Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸
Therefore no precipitate will form.
The answer that matches is option D
The value of Q has been less than the value of ksp. Thus, there will be no precipitate formed. Thus, option D is correct.
The precipitate has been the settling down of the compound with higher ksp. The ksp of a given calcium hydroxide solution has been [tex]8.0\;\times\; 10^-^8[/tex].
Computation for the precipitation:
The ksp for the solution of 0.064 g sample of calcium carbide has been:
[tex]\rm Q=ksp=[Ca]\;[OH]^-^2[/tex]
The concentration of calcium of 0.064 g sample in 1 L has been:
[tex]\rm Molarity=\frac{mass}{molar\;mass}\;\times\;\dfrac{1}{Volume\;L}[/tex]
The concentration of Calcium has been given by:
[tex]\rm Molarity\;Ca=\dfrac{0.064}{64}\;\times\;\dfrac{1}{1}\\Molarity\;Ca=0.001\;M[/tex]
The concentration of hydroxide ion for the reaction has been double the calcium ions.
Thus, the concentration of hydroxide ion has been:
[tex]\rm OH^-=2\;\times\;Ca^+\\OH^-=2\;\times\;0.001 \;M\\OH^-=0.002\;M[/tex]
The concentration of Calcium has been 0.001 M, and the concentration of hydroxide ion has been 0.002 M.
Substituting the values for the calculation of Q of the reaction.
[tex]\rm Q=[0.001]\;[0.002]^2\\Q=4.0\;\times\;10^-^9[/tex]
The value of Q for the reaction has been [tex]4.0\;\times\;10^-^9[/tex].
For the precipitation, the value of Q will be greater than the value of ksp. The value of Q has been less than the value of ksp. Thus, there will be no precipitate formed.
Thus, option D is correct.
For more information about ksp of reaction, refer to the link:
https://brainly.com/question/4736767
