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?

1 Answer
Jul 25, 2014

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