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:
Your answer is incorrect.
What is the experimental value of Ksp?
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.
Suppose a student titrates a 10.00-mL aliquot of saturated Ca(OH)2 solution to the equivalence point with...
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B. Ksp of Ca(OH)2 by titration
25 mL of Ca(OH)2 used***
Molarity of HCl solution = 0.0716M
1. Final buret reading (trial 1) = 7.50 mL
Final buret reading (trial 2) = 14.60 mL
2. Initial buret reading (trial 1) = 0.50 mL
Initial buret reading (trial 2) = 7.50 mL
3. Volume of HCl (trial 1) _________mL
Volume of HCl (trial 2) _________mL
4. Molarity of Ca(OH)2 (trial 1) __________M
Molarity of Ca(OH)2 (trial 2) ___________M
5. Average Molarity...
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