Chapt17Ksp_stu_10_ed_07

Chapt17Ksp_stu_10_ed_07 - Solubility Equilibria CuS(s)...

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Solubility Equilibria CuS(s) ”CuS(aq)” S Cu 2+ (aq) + S 2- (aq) 100 % dissociated usually written CuS(s) Cu 2+ (aq) + S 2- (aq ) K sp = [Cu 2+ ][S 2- ] K sp - solubility product constant
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Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH 1- (aq) K sp = [Mg 2+ ][OH 1- ] 2 BaCrO 4 (s) Ba 2+ (aq) + CrO 4 2- (aq) K sp = [Ba 2+ ][CrO 4 2- ] Ni 3 (PO 4 ) 2 (s) 3 Ni 2+ (aq) + 2 PO 4 3- (aq) K sp = [Ni 2+ ] 3 [PO 4 3- ] 2 X m Y n (s) m X n+ (aq) + n Y m- (aq) K sp = [X n+ ] m [Y m- ] n
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K sp holds for any slightly soluble salt can be determine from the molar solubility of the salt. MOLAR SOLUBILITY the concentration of the undissociated salt dissolved in a saturated solution. i.e. the [“CuS(aq)”] in the first example.
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Calculate the K sp of AgCl if its molar solubility is 9.6x10 –6 M .
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Calculate the K sp of Mg(OH) 2 if the molar solubility of Mg(OH) 2 = 1.5x10 –4 M .
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Calculate the molar solubility of ZnS. (K sp for ZnS = 1.0x10 –24 )
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Calculate the molar solubility of PbCl 2 . (K sp for PbCl 2 = 1.8x10 –4 )
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The presence of a common ion markedly reduce the solubility of a substance. markedly decreases the concentration of the other ion in a saturated solution.
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2 in 0.10 M NaOH. (K
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Chapt17Ksp_stu_10_ed_07 - Solubility Equilibria CuS(s)...

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