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:
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
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