Resolved Question

Calculate Ksp from titration experiment???10 points?
0.025L of Solution A [ Ca(OH)2 with 0.0125M of NaOH]...titrated against 0.0120L of HCl. [HCl] =0.1034
To break it down...
FIND: [OH-] from titration[OH-] from NaOH
[OH-] from Ca (OH)2 (by subtraction)[Ca2+]
Solubility of Ca(OH)2 in molL-1Ksp

by joeyeehu...

Best Answer - Chosen by Asker
No. of moles of HCl = 0.1034 x 0.0120 = 1.2408 x 10^-3 mol
That means total no. of moles of OH- = 1.2408 x 10^-3 mol
No. of moles of OH- from NaOH = 0.0125 x 0.025 = 3.125 x 10^-4
ThereforeNo. of moles of OH- from Ca(OH)2 = 1.2408 x 10^-3 - 3.125 x 10^-4= 9.283 x 10^-4
mol

Volume of solution A is 0.025L
Hence, [OH-] from Ca(OH)2 = 9.283 x 10^-4/ 0.025 M= 0.037132 M
[Ca2+] = 0.037132/2
Ksp = [Ca2+] [OH-]^2. = ( 0.037132) X { 0.037132/2 }^2. = 1.28 x 10^-5

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thanks