Question

When 0.277 mol of C4H9COOH is dissolved in 0.963 L of water, proton transfer occurs. C4H9COOH...

When 0.277 mol of C4H9COOH is dissolved in 0.963 L of water, proton transfer occurs.

C4H9COOH + H2O ⇄C4H9COO- + H3O+

The equilibrium concentration of H3O+ ions is 0.00212 M. Evaluate Keq.

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Answer #1

Let us write the reaction as

C4H9COOH + H2O <-------------> C4H9COO- + H3O+

0.277 55.5 0 0 initial moles

0.277 -x 55.5-x x x after reaction at equilibrium

Given at equilibrium [H+] = x = 0.00212 M  

thus moles of H+ in solution = molarity x volume

= 0.00212 x 0.963

=0.00204 moles

Thus

C4H9COOH + H2O <-------------> C4H9COO- + H3O+

0.277 55.5 0 0 initial moles

0.277 -x 55.5-x x x after reaction at equilibrium

0.277-0.00204 55.5-0.00204 0.00204 0.00204 moles at equilibrium

0.275/0.963 55.497/0.963 0.00204/0.963 0.00204/0.963 equilibrium concentrations

thus Keq = [C4H9COO-][H3O+]/ [C4H9COOH][H2O]

= 0.00204 x0.00204 /0.275 x 55.497

= 2.72 x10-7

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