The acid C6H5COOH dissociates according to the equation: C6H5COOH (aq) ⇌ C6H5COO‐ (aq) + H+ (aq)...

The acid C6H5COOH dissociates according to the equation: C6H5COOH (aq) ⇌ C6H5COO‐ (aq) + H+ (aq) K = 6.4 x 10‐5 If the initial [C6H5CO2H] is 4.59 x 10‐3M, what are the equilibrium concentrations of C6H5COO‐ and H+ for this dislocation reaction?

Solutions

Expert Solution

C6H5COOH (aq) <-------> C6H5COO‐ (aq) + H+ (aq)
4.59*10^-3 M 0 0
4.59*10^-3 -X X X

K = [C6H5COO‐] [H+] /[C6H5COOH ]
6.4*10^-5 = X*X / (4.59*10^-3 -X)
2.94*10^-7 - 6.4*10^-5* X = X^2
X^2 + 6.4*10^-5* X - 2.94*10^-7 = 0

Solving above quadratic equation we get,
X = 5.1*10^-4 M and X = -5.8*10^-4 M

But X can't be negative since it is concentration of product

So,
X = 5.1*10^-4 M

So equilibrium concetration of C6H5COO‐ and H+ is X which is equal to 5.1*10^-4 M

queen_honey_blossom answered 2 years ago

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