Molarity of HCl - 0.10000M Trial #1 Volume Ca(OH)2 - 10 ml Initial Buret Reading

Question

Molarity of HCl - 0.10000M

Trial #1

Volume Ca(OH)2 - 10 ml

Initial Buret Reading -0.10 mL

Final Buret Reading - 7.50 mL

Volume HCl Delivered - 7.40 mL

Solubility (M) - ?

Ksp of Ca(OH)2 - ?

Trail #2

Volume Ca(OH)2 - 10 ml

Initial Buret Reading - 0.50 mL

Final Buret Reading - 8.1 mL

Volume HCl Delivered - 7.6 mL

Solubility (M) - ?

Ksp of Ca(OH)2 - ?

QUESTIONS:

What is the average pKsp?

What is the percent error calculation for Average pKsp?

Using average Ksp predict the effect of 0.2M CaCl2 on molar solubility of Ca(OH)2. Create an ICE table for Ca(OH)2 setting initial conditions as having only CaCl2 in solution. Complete the ICE table with expressions for equilibrium concentrations and solve for solubility of Ca(OH)2

Explanation / Answer

Trial #1

moles of HCl = 0.1 M x 7.4 ml = 0.74 mmol

moles of Ca(OH)2 = 0.74/2 = 0.37 mmol

Solubility of Ca(OH)2 = 0.37 mmol/10 ml = 0.037 M

Ksp = [Ca2+][OH-]^2 = (0.037)(0.074)^2 = 2.0 x 10^-4

Trial #2

moles of HCl = 0.1 M x 7.6 ml = 0.76 mmol

moles of Ca(OH)2 = 0.74/2 = 0.38 mmol

Solubility of Ca(OH)2 = 0.38 mmol/10 ml = 0.038 M

Ksp = [Ca2+][OH-]^2 = (0.038)(0.076)^2 = 2.2 x 10^-4

Average Ksp = 2.1 x 10^-4

Percent error Ksp = (5.5 x 10^-6 - 2.1 x 10^-4) x 100/2.1 x 10^-4 = 2.14% (ignore - sign)

Molar solubility in presence of 0.2 M CaCl2

ICE table

            Ca(OH)2 <===> Ca2+     + 2OH-

initial         -                      0.2             -

change     -                        -x            +2x

Eq            -                      0.2-x          2x

x is a small change

2.1 x 10^-4 = (0.2)[2x]^2

molar solubility of Ca(OH)2 = sq.rt.(2.1 x 10^-4/4 x 0.2) = 0.016 M

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.