Benzoic acid reacts with water to form the
benzoate ion by the following reaction
C6H5COOH(aq) + H2O(l)
⇀↽ C6H5COO−(aq) + H3O
+(aq)
The equilibrium constant for this reaction is
6.4 x 10−5
. In a 0.1 M solution of benzoic acid,
what is the concentration of the benzoate ion
at equilibrium?
1. 0.0975 M
2. 2.5 x 10−3 M
3. 0.1 M
4. 6.4 x 10−5 M
5. 8.0 x 10−3 M
if benzoic acid ionize partially to x mole of both H30+ and C6H5COO-.At equilibrium, we have 0.1-xM of benzoic acid:6.4*10^-5=x^2/0.1-x since x is relatively small we have:x^2=6.4*10^-5*0.1 x=2.5*10^-3M
To solve this problem, let's assume the equilibrium concentration of the benzoate ion to be x M.
According to the balanced equation, the stoichiometric ratio between benzoic acid and benzoate ion is 1:1.
The equilibrium constant (Kc) for the reaction is given as 6.4 x 10^−5. This can be expressed as:
Kc = [C6H5COO−][H3O+]/[C6H5COOH]
Plugging in the values, we have:
6.4 x 10^−5 = x / (0.1 - x)
Solving this equation for x, we can find the concentration of the benzoate ion at equilibrium.
First, let's rearrange the equation:
6.4 x 10^−5 * (0.1 - x) = x
Multiplying both sides:
6.4 x 10^−6 - 6.4 x 10^−5x = x
Now, bring all the x terms to one side:
7.4 x 10^−5x + x = 6.4 x 10^−6
Combining like terms:
1.074 x 10^−4x = 6.4 x 10^−6
Dividing both sides by 1.074 x 10^−4:
x = 6.4 x 10^−6 / 1.074 x 10^−4
x ≈ 0.0596 M
Thus, the concentration of the benzoate ion at equilibrium in a 0.1 M solution of benzoic acid is approximately 0.0596 M.
Since none of the given answer choices match this result exactly, it's possible that there may be an error in the problem statement or calculation.
To find the concentration of the benzoate ion (C6H5COO-) at equilibrium, we can use the equilibrium constant (Kc) and the initial concentration of benzoic acid (C6H5COOH). The balanced chemical equation for the reaction is:
C6H5COOH(aq) + H2O(l) ⇀↽ C6H5COO-(aq) + H3O+(aq)
The equilibrium constant expression for this reaction is given by:
Kc = [C6H5COO-][H3O+]/[C6H5COOH][H2O]
The concentration of water is generally considered constant, so we can ignore it in the equilibrium constant expression:
Kc = [C6H5COO-][H3O+]/[C6H5COOH]
Given that the equilibrium constant (Kc) is 6.4 x 10^-5 and the initial concentration of benzoic acid is 0.1 M, we can set up the following expression:
6.4 x 10^-5 = [C6H5COO-][H3O+]/0.1
Since the reaction stoichiometry is 1:1 between benzoic acid and benzoate ion, the concentration of benzoate ion at equilibrium ([C6H5COO-]) will be equal to the concentration of H3O+ at equilibrium ([H3O+]). Let x be the concentration of benzoate ion at equilibrium, so we have:
6.4 x 10^-5 = x^2/0.1
Simplifying the equation, we obtain:
x^2 = 6.4 x 10^-5 * 0.1
x^2 = 6.4 x 10^-6
Taking the square root of both sides, we find:
x = sqrt(6.4 x 10^-6)
x ≈ 2.531 x 10^-3 M
Therefore, the concentration of the benzoate ion at equilibrium in a 0.1 M solution of benzoic acid is approximately 2.531 x 10^-3 M.
Therefore, the correct answer is option 2: 2.5 x 10^-3 M.