Question

How do you calculate the pH of a weak acid?

Answer 1

A handy expression to use is `pH=1/2(pK_a-loga)`

Explanation:

This is a quick and useful method if you are given the `pK_a` and concentration `a` of the weak acid.

I'll derive it using the usual ICE method:

`HX_((aq))rightleftharpoonsH_((aq))^(+)+X_((aq))^(-)`

`"Initial:"` `" "`a`" "" "0" " " " " "0`

If `x` moles dissociate the equilibrium moles `rArr`

`(a-x) " " " " " " " " x " " " " " " x`

`K_a=(x^2)/(a-x)`

We assume that `x` is negligible compared to to `a` so this approximates to:

`K_a=(x^2)/(a)`

`(x^2)/(a)=K_a`

`x^2=K_axxa`

`x=(K_axxa)^((1)/(2))`

Taking -ve logs of both sides `rArr`

`-logx=-1/2log(K_a+a)`

`pH=1/2(pK_a-loga)`

Here's an example:

`pH` of a `0.1"M"` solution of ethanoic acid whose `pK_a=4.75` ?

`pH=1/2(4.75-(-1))`

`pH=5.75`

If you can remember it then fine. If not you need to use the ICE method described in the answer by @Dr Hayek

Answer 2

`pH = -log[H_3O^+]`

Explanation:

By measuring the concentration of `H_3O^+` ions we get the pH using `pH = -log[H_3O^+]`

Consider the dissociation of the following weak acid HA with initial concentration of `1.0 M` and `K_a = 2.0 xx 10^-6`:

`" " " " " " " " " " "` `HA(aq) + H_2O(aq) -> H_3O^+(aq) + A^(-)(aq)`
`" " " " " "` Initial: `1.0 M " " " " " " " " " " " " 0 M " " " " " " 0M`
`" " " " "` Change: `- xM " " " " " " " " " " " +x M" " " " " +x M`
`" " "` Equilibrium: `(1.0- x)M " " " " " " " " "x M" " " " " " " " x M`

The expression of `K_a` can be written as follows:

`K_a = ([H_3O^+(aq)]*[A^(-)(aq)])/([HA(aq)])=(x*x)/(1.0-x)=2.0xx10^-6`

solve for `x` using calculator, you get

`x~~1.4xx10^-3 => pH = -log(1.4xx10^-3) = 2.895`

that can be rounded to `pH=2.9` (2 significant figures).

Here is a video which summarizes this technique which requires use of the quadratic formula. The values are different than Dr. Hayek's example, but the technique is the same.

Video from: Noel Pauller