Question

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 7.0×10−2 M .

Answer 1

To determine the equilibrium concentration of H3O+ for benzoic acid with a given initial concentration and acid-dissociation constant (Ka), set up an ICE table and solve for the variable representing the change in concentration, which is also the concentration of H3O+ at equilibrium.

Step-by-step explanation:

The acid-dissociation constant (Ka) is used to calculate the equilibrium concentration of H3O+ for a weak acid like benzoic acid (C6H5COOH). To find the concentration of H3O+ at equilibrium when the initial concentration of C6H5COOH is 7.0×10⁻² M and the Ka is 6.3×10⁻⁵, we can set up an ICE table (Initial, Change, Equilibrium) to solve for the change in concentration (x) that occurs as the acid dissociates.

Initial concentrations are [C6H5COOH] = 7.0×10⁻² M, [C6H5COO⁻] = 0, [H3O+] = 0. At equilibrium, the concentrations will be [C6H5COOH] = 7.0×10⁻² - x, [C6H5COO⁻] = x, [H3O+] = x. Since Ka = [C6H5COO⁻][H3O+]/[C6H5COOH], we can substitute the equilibrium values and solve for x, which represents the equilibrium concentration of H3O+.

Assuming x is small compared to the initial concentration, the equation simplifies to Ka ≈ x²/(7.0×10⁻²), which can be solved for x to give the concentration of H3O+.

Answer 2

Answer : The equilibrium concentration of
`H_3O^+` in the solution is,
`2.1* 10^(-3)M`

Explanation :

The dissociation of acid reaction is:

`C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-`

Initial conc. c 0 0

At eqm. c-x x x

Given:

c =
`7.0* 10^(-2)M`

`K_a=6.3* 10^(-5)`

The expression of dissociation constant of acid is:

`K_a=([H_3O^+][C_6H_5COO^-])/([C_6H_5COOH])`

`K_a=((x)* (x))/((c-x))`

Now put all the given values in this expression, we get:

`6.3* 10^(-5)=((x)* (x))/([(7.0* 10^(-2))-x])`

`x=2.1* 10^(-3)M`

Thus, the equilibrium concentration of
`H_3O^+` in the solution is,
`2.1* 10^(-3)M`