Ca(OH)2 ksp的計算 (急,,15點)

No. of moles of H+ ions added = 0.123 x 0.02655 = 0.003266

Hence no. of moles of OH- ions present in the solution = 0.003266

in which 0.1229 x 0.01 = 0.001229 moles are from NaOH

Hence no. of moles of OH- ions from Ca(OH)2 = 0.003266 - 0.001229 = 0.002037

So No. of moles of aqueous Ca2+ ions = 0.002037/2 = 0.001018

Thus, before any titration occurs:

[Ca2+] = 0.001018/0.1 = 0.01018

[OH-] = 0.003266/0.1 = 0.03266

Ksp of Ca(OH)2 = 0.01018 x 0.032662 = 1.086 x 10-5 M3

2010-11-29 08:58:31 補充：
Moreover, it should be noted that the Ca(OH)2 has solid residues undissolved, that is:

No. of moles of Ca2+ ions = 0.001018

Hence no. of moles of Ca(OH)2 dissolved = 0.001018, with mass = 0.0753 g

As mass of Ca(OH)2 originally present = 2.1177 g, it has residues undissolved.

2010-11-29 15:18:06 補充：
I think the question should have mentioned "10 cm^3 of NaOH" rather than "100 cm^3".

問題的數字有錯?
假設Ca(OH)2 完全不溶, 即是所有OH(-) 都來自NaOH;
no. of mole of NaOH = 0.1229 x (100/1000) = 0.01229 mole
no. of mole of HCl = 0.123 x (26.55/1000) = 0.00326565 mole
HCl 的mole 數比NaOH 的mole 數小很多, 根本不可能用那些許的酸就滴定出來.

? 酸的濃度太低 /
? 酸的volume 太小 /
? NaOH 的濃度太高 /
? NaOH 的volume 太大.