Question

A 1.0 M solution of a weak acid, HB, has an [H3O+] equal to 0.025 M. Find the value of Ka. (HB + H2O --> B- + H3O+) Can anyone help?

Answer 1

`K_text(a) = 6.4 × 10^"-4"`

Explanation:

We can use an ICE table to help us solve this problem.

`color(white)(mmmmmmll)"HB" + "H"_2"O" ⇌ "H"_3"O"^"+" + color(white)(l)"B"^"-"`
`"I/mol·L"^"-1": color(white)(mll)1.0 color(white)(mmmmmmll)0color(white)(mmml)0`
`"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x`
`"E/mol·L"^"-1": color(white)(m)"1.0-"xcolor(white)(mmmmmm)xcolor(white)(mmml)x`

At equilibrium,

`["H"_3"O"^"+"] = "0.025 mol/L"`

∴ `x = 0.025`

`K_text(a) = (["H"_3"O"^"+"]["B"^"-"])/(["HB"]) = x^2/(1.0-x) = 0.025^2/(1.0-0.025) = (6.25 × 10^"-4")/0.975 = 6.4 × 10^"-4"`