Question

0.02500 M EDTA solution were used to titrate 10.00 mL saturated Ca(OH)2 solution. EDTA=C10H14N2Na2O8 The average volume of EDTA solution needed was 11.15 mL to titrate a 10.00 mL saturated solution of Ca(OH)2.

milimoles of EDTA= 0.02500(M) *11.15ml = 0.2875milimoles

Milimoles of Ca2+= strength * volume = 2.845milimoles

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1.) How many grams of Ca(OH)2 can be dissolved in 1-L solution?

2.) Ksp

Answer 1

Answer:

EDTA

concentration (M₁) = 0.025 M Volume (V₁) = 11.15 mL

Ca(OH)2

M₂ = ? Volume (V₂) = 10 mL

From the Molarity principle M₁V₁ = M₂V₂

0.025 x 11.15 = M₂ x 10

10 x M₂ = 0.27875

M₂ = 0.27875 / 10

M₂ = 0.02787 M

Ca(OH)₂ weight caluculation:

Molarity = Number of moles / Volume in Liter

0.02787 = Number of moles / 1

Number of moles = 0.02787

moles = weight / molar mass

molar mass of Ca(OH)₂ = 74

0.02787 = weight / 74

weight = 0.02787 x 74

weight = 2.06238 g

2) Ksp of Ca(OH)₂ caluculation:

Ca(OH)₂   Ca²⁺ + 2 ^-OH

solubility of Ions= S ........2S

For Ca²⁺ is S and for 2 hydroxyl ions the solubility value is 2S

Solubility product Ksp = [S].[2S]²

Ksp = S . 4S²

Ksp = 4S³

S = solubilty = concentration of the Ca(OH)₂

Ksp = 4 x (0.02787)³

Ksp = 4 x (0.02787)³

Ksp = 4 x 21.647 x 10^-6

Ksp = 86.588 x 10^-6 moles³ / liter³