Question

What is the original molarity of a solution of a weak acid, given Ka and pH?

`K_a=3.5*10^-5`
`pH=5.25`
@25 degrees Celsius

Answer 1

`6.5*10^-6` M

Explanation:

Construct an ICE table using the following reaction equation:
`H_2O` + `HA` `rightleftharpoons` `A^-` + `H_3O^+`

Use the pH to calculate `[H_3O^+]` at equilibrium, which is also the change in concentration for the table.

Equilibrium Concentrations:
`[HA] = x-5.6*10^-6` M
`[A^-]=5.6*10^-6` M
`[H_3O^+]=5.6*10^-6` M

Set up an equilibrium expression using `K_a`:
`3.5*10^-5=(5.6*10^-6)^2/(x-5.6*10^-6)`
`x=9.0*10^-7`

`[HA]=9.0*10^-7` M `-5.6*10^-6` M `=6.5*10^-6` M