Question

Ions B and C to form the complex BC. If 35.0 mL of 1.00 M B is combined with 35.0 mL of 1.00 M C, 0.00500 mol of BC is formed. Determine the equilibrium constant fo this reaction.

Answer 1

Given, 35.0 mL of 1.00 M B.

So, moles of B = Molarity x Volume(in L) = 1.00 M x 0.035 L = 0.035 mol

Total volume = 35.0 mL of B + 35.0 mL of C = 70.0 mL

[B] = moles / Total volume = 0.035mol / 0.070 L = 0.5 M

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Given, 35.0 mL of 1.00 M C.

So, moles of C = Molarity x Volume(in L) = 1.00 M x 0.035 L = 0.035 mol

Total volume = 35.0 mL of B + 35.0 mL of C = 70.0 mL

[C] = moles / Total volume = 0.035mol / 0.070 L = 0.5 M

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Given, 0.00500 mol of BC.

Total volume = 35.0 mL of B + 35.0 mL of C = 70.0 mL

[BC] = moles / Total volume = 0.00500mol / 0.070 L = 0.071 M

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Thus,

[B] = 0.5 M

[C] = 0.5 M

[BC] = 0.071 M

             B      +      C        <------------->          BC

I           0.5             0.5                                  0

C          -x                -x                                  +x

E         0.5 - x         0.5-x                              0.071

So, [BC] = x = 0.071

[B] = 0.5-x = 0.5 - 0.071 = 0.43 M

[C] = 0.5-x = 0.5-0.071 = 0.43 M

Kc = [BC] / [B][C]

Kc = [0.071] / [0.43]^2

Kc = 0.384