Question

In: Chemistry

Find the equilibrium concentration of the Mn(C2O4)2− ion(in M) when 0.50 L of 0.030 M Mn(NO3)2...

Find the equilibrium concentration of the Mn(C2O4)2− ion(in M) when 0.50 L of 0.030 M Mn(NO3)2 are mixed with 1.0 L of 1.2M Na2C2O4. Kf of Mn(C2O4)2 2- is 6.3 x 10 5.

Expert Solution

moles Mn2+ = 0.03 M x 0.5 L = 0.015 mmol

moles C2O4^2- = 1.2 M x 1.0 L = 1.2 mmol

ICE chart

Mn2+ + 2C2O4^2- <===> [Mn(C2O4)2]^2-

I 0.015 1.2 -

C -x -2x +x

E 0.015-x 1.2-2x x

So,

Kf = [Mn(C2O4)2]2-/[Mn2+][C2O4^2-]^2

6.3 x 10^5 = x/(0.015-x)(1.2-2x)^2

1.44 - 4.8x + 4x^2

0.0216 - 0.072x + 0.06x^2 - 1.44x + 4.8x^2 - 4x^3

-4x^3 + 4.86x^2 - 1.512x + 0.0216

-2.52 x 10^6 x^3 + 3.062 x 10^6 x^2 - 9.52 x 10^5 x + 1.361 x 10^4 = 0

x = 0.015 mmol

So,

equilibrium concentration of [Mn(C2O4)2]^2- = 0.015 mmol/1.5 L = 0.01 M

queen_honey_blossom answered 2 years ago

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