Question

A saturated solution of the Ca(Oh)₂ is pepared and allowed to equilibrate at 25^oC. A 25 mL sample of the solution is titrated with 0.0050M HCl solution. 1.4mL of the HCl solution are required to fully react with the sample. a) Write a balanced chemical equation of the salt with HCl. b)Calculate the molarity of the salt solution from the titration data. c) Calculate the concentration of the cation and the anion in the salt solution. d)Show the calculation for the K_SP of the salt from the concentration of the cation and anion in the salt solution. E) Calculate the deltaG fot the salt using the K_SP you calculated for the question above. f) Is the reation sponteneous or nonsponteneous? Why did you answer that way? Of you calculate the deltaH for this reaction, what does that tell you about your reaction?

Answer 1

a) The balanced chemical equation of Ca(OH)₂ with HCl is;

Ca(OH)₂ + 2HCl CaCl₂ + 2H₂O

b) The molarity of the salt solution from the titration data, is as follows

1.4mL of 0.0050M HCl solution.

Moles of HCl = Molarity x volume (L)

                   = 0.0050 M x 0.0014 L

                   = 0.0000070 moles

Since 1 mole of HCl reacts with 0.5 moles of Ca(OH)₂.

so, 0.0000070 moles of HCl reacts with (0.5 x 0.0000070 = 0.0000035) moles of Ca(OH)₂.

So, moles of Ca(OH)₂ = 0.0000035

Total volume = 25 mL + 1.4 mL = 26.4 mL = 0.0264 L

So, molarity of Ca(OH)₂ = 0.0000035 moles / 0.0264 L = 0.000133 M

C)

Ca(OH)₂ + 2HCl    CaCl₂ + 2H₂O

After reaction CaCl₂ is present

CaCl₂ Ca²⁺ + 2Cl^-

0.0000035 moles of Ca(OH)₂ will produce 0.0000035 moles mole of Ca(OH)₂. But the total volume is 26.4 mL

So, molarity of Ca(OH)₂ = molarity of CaCl₂ = 0.000133 M

So,

[Ca²⁺] =0.000133 M and

[Cl^-] = (2 x 0.000133) M = 0.000266 M

d)

Ca(OH)₂ Ca²⁺ + OH^-

molarity of Ca(OH)₂ = 0.000133 M

Ksp = [Ca²⁺] [OH^-]²

       = (0.000133) x (0.000266)²

       = (1.33 x 10^-4 ) x (2.66 x 10^-4 )²

       = 3.54 x 10^-12