Question

Create an ICE table and substitute the experimental Ka value (2.69*10^-4) to determine the pH of a 0.0020 M solution.

Requires use of the quadratic formula.

Answer 1

lets your acid will be HA

HA + H₂O ----> H₃O⁺ + A^-

I 0.002 0 0

C -x +x +x

E 0.002-x +x +x

dissociation constant

Ka = [H₃O⁺][A^-] / [HA]

2.69 *10^-4 = [x][x] / [0.002-x]

2.69 *10^-4 [0.002-x] = x²

x² + x2.69 *10^-4 - 5.38 * 10^-7 =0 this is the quadratic equation

i am using the online sodt wear to calculate the quadratic equation

i got the two values for x one is in +ve and another one is in -Ve concentration cannot be a -ve value so i am taking positive value

x = 0.0006112 M = [H3O+] = [A-]

now you have the concentration of H3O+ from this we can find out pH

pH = -log[H3O+]

pH = -log[0.0006112]

pH = 3.214