Question

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

Answer 1

The concentration of copper in the solution is

$5.1 \times 10^{18} \: M.$

The volume of the solution = 1.00 L

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

$Moles \: of \: Ethylenediamine = 1.20 \times 10.3$

$k _{f} = 1.00 \times 10^{20}$

Overall balanced equation of reaction is,

$Cu^{2 + }(aq) \: + 2en \: (aq)→ Cu(en) ^{2 + }_{2}(aq)$

$Moles \: of \:Cu(NO_{3}) _{2} =0.00025 \: mol$

Mole ratio for,

$Moles \: of\:Cu(NO_{3}) _{2}: en$

$2: 1$

$Moles \: of \: en = 0.000250 \times 2$

$= 0.00050 \: moles$

Remained moles of en are= (0.00120-0.000500)

$= 0.000700 \: moles$

The concentration of copper in the solution is,

$k_{f} = \frac{Cu(en)^{2 + }_{2} }{Cu^{2+}en^{2} }$

$1.00 \times 10 ^{20} = \frac{0.000250}{(Cu )^{2 +} \times (0.000700)^{2 + } }$

$= 5.1 \times 10^{18} \: M.$

Therefore, the concentration of copper in the solution is

$5.1 \times 10^{18} \: M.$

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

brainly.com/question/8732513

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