Question

The `K_"a"` for the weak acid HA is `4.0xx10^-6`. what is the pH of a `0.01"M"` solution of the acid? what is its `"p"K_"a"`?

Answer 1

`"p"K_"a" = 5.4`

`"pH" = 3.7`

Explanation:

`"p"K_"a"` is simply `-log_10(K_"a")`. It is a constant at any temperature and does not depend on the molarity of the solution.

`"p"K_"a" = -log_10(K_"a")`

`= -log_10(4.0xx10^{-6})`

`= 5.4`

To find the pH of the solution, we let the extent of dissociation be `x`.

Initially, before dissociation occurs, the concentration of each species, in `"M"`, are as follows.

• `["HA"] = 0.01`
• `["H"^+] = 0`
• `["A"^-] = 0`

After `x` amount of dissociation has occurred, the solution is in equilibrium. The concentration of each species, in `"M"`, are as follows.

• `["HA"] = 0.01-x`
• `["H"^+] = x`
• `["A"^-] = x`

Since they are in equilibrium, the relation below holds

`K_"a" = frac{["H"^+] * ["A"^-]}{["HA"]}`

`4.0xx10^{-6} = frac{x * x}{0.01-x}`

This is a quadratic equation. But note that since `K_"a"` is small, the acid is a weak acid. Therefore, dissociation occurs to a little extent only. Hence, we can assume `x "<<" 0.01` to simplify our calculations.

`4.0xx10^{-6} = frac{x^2}{0.01-x}`

`~~ frac{x^2}{0.01}`

Thus,

`x = sqrt{(4.0xx10^{-6}) xx 0.01}`

`= 2.0 xx 10^{-4}`

Our original assumption of `x "<<" 0.01` is valid and we can check that

`frac{(2.0 xx 10^{-4})^2}{0.01-(2.0 xx 10^{-4})} = 4.008 xx 10^{-6}`

`~~ 4.0 xx 10^{-6}`

Since we know `x`, we know that at equilibrium,

`["H"^+] = x = 2.0 xx 10^{-4}`

The pH of the solution is given by

`"pH" = -log_10(["H"^+])`

`= -log_10(2.0 xx 10^{-4})`

`= 3.7`