Question

A 0.100 M solution of an acid, HA, has a pH = 2.00. what is the value of the ionization constant, `K_a` for this acid?

Answer 1

`K_"a" = 0.11`

Explanation:

Let's set up an ICE table to solve this problem.

`color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"`
`"I/mol·L"^"-1":color(white)(ml)0.100color(white)(mmmmml)0color(white)(mmm)0`
`"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)+xcolor(white)(ml)+x`
`"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmll)xcolor(white)(mmll)x`

We must use the `"pH"` to calculate the value of `x`.

`"pH = 2.00"`

`["H"_3"O"^"+"] = 10^"-2.00" color(white)(l)"mol/L" = "0.0100 mol/L" = x`

The `K_"a"` expression is:

`K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x)`

∴ `K_"a" = x^2/(0.100-x) = 0.100^2/(0.100-0.010) = 0.0100/0.090 =0.11`