Answered

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At equilibrium, 1.0 mol of a 4.5 x 10-4 M solution of substance A reacts with 1 mol of a solid to form 2.0 mol of a 1.2 x 10-2 M solution of substance C and 1.0 mol of a solution of substance D. Given that K = 2.0 x 10-6, what is the equilibrium concentration of substance D?

Sagot :

Step 1

The reaction:

1 A (aq) + 1 B (s) <=> 2 C (aq) + 1 D (aq)

A, C and D participate in the equilibrium constant K

B doesn't participate

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Step 2

Data provided:

At equilibrium:

[A] =4.5 x 10^-4 M

[C] = 1.2 x 10^-2 M

[D] = Unknown

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Step 3

[tex]K\text{ = }\frac{\lbrack D\rbrack^1\lbrack C\rbrack^2}{\lbrack A\rbrack^1}=\frac{\lbrack D\rbrack.(1.2x10^{-2}M)^2}{(4.5x10^{-4}M)}=\text{ 2.0 x 10}^{-6}[/tex]

Answer:

[D] = 6.25x10^-6 M

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