Iminodiacetic acid forms 2:1 complexes with many metal ions: A 25.0 mL solution containing 0.120 M iminodiacetic
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Iminodiacetic acid forms 2:1 complexes with many metal ions:
A 25.0 mL solution containing 0.120 M iminodiacetic acid buffered to pH 7.00 was titrated with 25.0 mL of 0.050 0 M Cu2+.Given that αx2 = 4.6 × 10-3 at pH 7.00, calculate [Cu2+] in the resulting solution.
Transcribed Image Text:
CH,CO,H H,Ñ CH,CO,H H,X* [X?-1 ax? [H;X*]+ [H2X] + [HX¯] + [X ] Cu²+ + 2x?- = CuX K = B2 = 3.5 × 1016
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Given Data Volume 25 mL pH 700 Initial moles Cu2 present 005 M x 25 ml 125 mmol initial H ...View the full answer
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