Calculate the equilibrium concentration of H3O+ in the solution if the initial concentration of C6H5COOH is 6.0×10^-2. Kc=6.3 *10^-5
To calculate the equilibrium concentration of H3O+ in the solution, we need to use the equilibrium constant expression (Kc) and set up an ICE (Initial - Change - Equilibrium) table.
Given the equilibrium constant (Kc = 6.3 * 10^-5) and the initial concentration of C6H5COOH (6.0 × 10^-2), we can assume that the initial concentration of H3O+ is zero since it is a product of the reaction.
The balanced equation for the dissociation of C6H5COOH is: C6H5COOH ⇌ C6H5COO- + H3O+
Now let's set up the ICE table:
C6H5COOH ⇌ C6H5COO- + H3O+
I) 6.0 × 10^-2 0 0
C) -x x x
E) 6.0 × 10^-2 - x x x
Since the initial concentration of H3O+ is zero and it is produced in the reaction, its concentration at equilibrium will be equal to x.
According to the equilibrium expression of Kc, we have:
Kc = [C6H5COO-] · [H3O+] / [C6H5COOH]
Substituting the values into the expression:
6.3 × 10^-5 = (x) · (x) / (6.0 × 10^-2 - x)
Now we need to solve this equation for x. Rearrange the equation to isolate x:
6.3 × 10^-5 = x^2 / (6.0 × 10^-2 - x)
Multiply both sides by (6.0 × 10^-2 - x):
(6.0 × 10^-2 - x) · (6.3 × 10^-5) = x^2
Expand and rearrange:
3.78 × 10^-6 - 6.3 × 10^-5x = x^2
Rearrange the equation to form a quadratic equation:
x^2 + 6.3 × 10^-5x - 3.78 × 10^-6 = 0
Now we can solve this quadratic equation using the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / (2a)
Plugging in the values: a = 1, b = 6.3 × 10^-5, and c = -3.78 × 10^-6:
x = [-6.3 × 10^-5 ± √((6.3 × 10^-5)^2 - 4(1)(-3.78 × 10^-6))] / (2(1))
Simplifying:
x = [-6.3 × 10^-5 ± √(3.969 × 10^-9 + 1.512 × 10^-5)] / 2
x = [-6.3 × 10^-5 ± √(1.512003969 × 10^-5)] / 2
x = [-6.3 × 10^-5 ± 1.229708445 × 10^-3] / 2
There are two possible solutions for x:
1. x = (-6.3 × 10^-5 + 1.229708445 × 10^-3) / 2 = 5.932 × 10^-4
2. x = (-6.3 × 10^-5 - 1.229708445 × 10^-3) / 2 = -6.285 × 10^-4
Since concentrations cannot be negative, we discard the second solution. Therefore, the equilibrium concentration of H3O+ in the solution is approximately 5.932 × 10^-4.
This is benzoic acid; let's call it HBz
..............HBz ==> H^+ + Bz^-
initial......0.06.....0......0
change........-x.......x.....x
equil.......0.06-x.....x.....x
K + (H^+)(Bz^-)/(HBz)
Substitute and solve for (H^+).