Question

A 0.20 mol/L solution of propionic acid has a pH of 2.79. What is the `"p"K_text(a)` of propionic acid?

Answer 1

The `"p"K_"a"` of propanoic acid is 4.88.

First, calculate the equilibrium concentration of H₃O⁺.

pH = 2.79

[H₃O⁺] = `10^"-pH"` mol/L= `10^-2.79` mol/L = 1.62 × 10⁻³ mol/L

Second, write the balanced chemical equation and set up an ICE table.

HA +H₂O ⇌ A⁻ + H₃O⁺
I/mol·L⁻¹: 0.20; 0; 0
C/mol·L⁻¹: `-x`; `+x`; `+x`
E/mol·L⁻¹: 0.20 - `x`; `x`; `x`

We know that at equilibrium, [H₃O⁺] = [A⁻] = `x` = 1.62 × 10⁻³ mol/L

and [HA] = (0.20 - `x`) mol/L = (0.20 - 1.62 × 10⁻³) mol/L = 0.198 mol/L

Third, calculate `K"a"`.

`K"a" = ([H₃O⁺][A⁻])/([HA]) =(1.62 × 10⁻³ × 1.62 × 10⁻³) /0.198` = 1.32 × 10⁻⁵

Fourth, calculate `"p"K_"a"`

`"p"K_"a" = -logK_"a"` = log(1.32 × 10⁻⁵) = 4.88