Question

Trial	Precise volume of Ca(OH)2 solution	Equivalence point Volume of HCl (mL)
1	14 mL	11 mL

1.Calculate the OH^- in the saturated Ca(OH)₂ sol. from the results of this titration

2.Calculate the Ca²⁺

3. Calculate the Ksp for Ca(OH)₂

Answer 1

I think 0.039 M will be molarity of HCl. Because concentration of unknown compunds will determined by titrating wih solutions of known concentration.

Given that [Ca(OH)₂] = 0.039 M

volume of Ca(OH)₂ = 14 mL = 0.014 L

Therefore, concentration of Ca(OH)2 = molarity / volume = 0.039/0.014 = 2.785 moles

Ca(OH)₂ ---------------> Ca⁺² + 2OH^-

   2.785 moles Ca(OH)₂ ---------------> 2.785 moles Ca⁺² + 2.785 moles x 2OH^-

   [OH-] = 2.785 moles x 2 = 5.57 moles

[Ca⁺²]=2.785 moles

Ksp for Ca(OH)₂

   Ca(OH)₂ ---------------> Ca⁺² + 2 OH^-

Ksp =   [Ca⁺²] [2 OH^-]²

= 4 [Ca⁺²] [ OH^-]²

= 4 (2.785 moles ) ( 5.57 moles )²

= 345.61

Ksp for Ca(OH)₂ = 345.61

( Generally Ksp will be very small)

I think 0.039 M will be molarity of HCl. Because concentration of unknown compunds will determined by titrating wih solutions of known concentration.

Ca(OH)2 + 2HCl -------------> CaCl2 + 2H2O

Given that

[Ca(OH)2] , M1= ?

Volume of Ca(OH)2, V1 = 14 mL = 0.014 L

Moles of Ca(OH)2, n1= 1 mol

[HCl], M2 = 0.039 M

volume of HCl , V2= 11 mL = 0.011 L

  Moles of Ca(OH)2, n2= 2 mol

At equivalence point , M1V1/n1 = M2V2/n2

M1X0.014L/1 = 0.039Mx 0.011 L/2

M1 = 0.0153 M

Therefore, concentration of Ca(OH)₂ = 0.0153 M

Ca(OH)₂ ---------------> Ca⁺² + 2OH^-

   0.0153 M  Ca(OH)₂ --------------->  0.0153 M Ca⁺² +  0.0153 M x 2OH^-

     [OH-] =  0.0153 M x 2 = 0.0306 M

[Ca⁺²]= 0.0153 M

Ksp for Ca(OH)₂

   Ca(OH)₂ ---------------> Ca⁺² + 2 OH^-

Ksp =    [Ca⁺²] [2 OH^-]²     

= 4  [Ca⁺²] [ OH^-]²

= 4 (0.0153 ) ( 0.0306 )²

= 5.73 x10^-5

Ksp for Ca(OH)₂ =  5.73 x10^-5

Therefore,

[OH-] =  0.0153 M x 2 = 0.0306 M

[Ca⁺²]= 0.0153 M

  Ksp for Ca(OH)₂ =  5.73 x10^-5