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Suppose a student titrates a 10.00-mL aliquot of saturated Ca(OH)2 solution to the equivalence point with...

Suppose a student titrates a 10.00-mL aliquot of saturated Ca(OH)2 solution to the equivalence point with 14.30 mL of 0.0224 M HCl. What was the initial [OH − ]? WebAssign will check your answer for the correct number of significant figures

.064 M 1 Incorrect:

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What is the experimental value of Ksp?

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Answer #1

Since 1 mole of Ca(OH)2 is completely neutralized by 2 mole of HCl

so, 2X moles of HCl = mole of Ca(OH)2

i.e. 2X14.30 mL X 0.0224 mol /L = 10.00 mLX Molarity (mole/L)

this gives , Molarity = 2X 14.30 X0.0224/10.00 = 0.0641 M

Ksp of CaCl2 can be calculated as: please see the uploaded picture.

Cle Colo Car (ar) * 2 ce(ar) DATE: Cla DATE : / PAGE NO.: - Cachos Ksp = [cat] [cé (a)]² = S. (20%= 48 kop as s= 0.0641M so,

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