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:
![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](https://dsd5zvtm8ll6.cloudfront.net/si.question.images/images/question_images/1591/8/7/4/0055ee211d5cd1f31591874005099.jpg)
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.
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To calculate the Cu2 in the resulting solution we need to use the equation for the formation const...View the full answer
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