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a 1.00-l solution contains 2.50×10-4 m cu(no3)2 and 1.20×10-3 m ethylenediamine (en). the kf for cu(en)22 is 1.00 × 1020. chegg

Sagot :

The concentration of copper in the solution is

[tex]5.1 \times 10^{18} \: M.[/tex]

The volume of the solution = 1.00 L

[tex]Moles \: of \:Cu(NO_{3}) _{2} = 2.50 \times 10 {}^{ - 4} [/tex]

[tex]Moles \: of \: Ethylenediamine = 1.20 \times 10.3[/tex]

[tex]k _{f} = 1.00 \times 10^{20} [/tex]

Overall balanced equation of reaction is,

[tex]Cu^{2 + }(aq) \: + 2en \: (aq)→ Cu(en) ^{2 + }_{2}(aq)[/tex]

[tex]Moles \: of \:Cu(NO_{3}) _{2} =0.00025 \: mol[/tex]

Mole ratio for,

[tex]Moles \: of\:Cu(NO_{3}) _{2}: en[/tex]

[tex]2: 1[/tex]

[tex]Moles \: of \: en = 0.000250 \times 2[/tex]

[tex] = 0.00050 \: moles[/tex]

Remained moles of en are= (0.00120-0.000500)

[tex] = 0.000700 \: moles[/tex]

The concentration of copper in the solution is,

[tex]k_{f} = \frac{Cu(en)^{2 + }_{2} }{Cu^{2+}en^{2} }[/tex]

[tex]1.00 \times 10 ^{20} = \frac{0.000250}{(Cu )^{2 +} \times (0.000700)^{2 + } }[/tex]

[tex] = 5.1 \times 10^{18} \: M.[/tex]

Therefore, the concentration of copper in the solution is

[tex]5.1 \times 10^{18} \: M.[/tex]

To know more about molarity, refer to the below link:

https://brainly.com/question/8732513

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