Question

Ina solution prepared by dissolving .100 mole of pro panic acid in enough water to make 1L of solution, the ph is observed to be 2.924. The Ka for propanoic acid (HC3H5O2) is

Answer 1

Answer:

The Ka for propanoic acid (HC3H5O2) is 2.1 x 10⁻⁵

Explanation:

Concentration of propanoic acid C = moles / volume in Litres = 0.1 mol / 1 L  = 0.1 M

C = 0.1 M

pH = - log [H+]

For weak acids,

[H+] = $Ka.C^{1/2}$

pH = - log ($Ka.C^{1/2}$)

$Ka.C^{1/2}$ =  $10^{-PH}$

Ka.C = ($10^{-PH}$)²

Ka = ($10^{-PH}$)² / C

     = ($10^{-2.832}$)² / 0.1

      = 2.1 x 10⁻⁵

Therefore,

Ka of propanoic acid = 2.1 x 10⁻⁵