Question

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In the laboratory, a general chemistry student measured the pH of a 0.550 M aqueous solution...

In the laboratory, a general chemistry student measured the pH of a 0.550 M aqueous solution of formic acid, HCOOH to be 1.987.

Use the information she obtained to determine the K_a for this acid.

K_a(experiment) =

Expert Solution

we have to use ICE table to calculate this

HCOOH + H2O <-------> HCOO^- + H3O⁺

dissociation constant Ka expression for this equation

Ka = [HCOO-] [H3O+] / [HCOOH]

ICE [ HCOOH ] [ HCOO^- ] [ H3O⁺ ]

initial concentration 0.55 0 0

Change in concentration x -x -x

equilibrium concentration 0.55 - x x x

x = [H3O⁺]

pH = -log[H3O+]

pH he has given 1.987

1.987 = -log[H3O+]

[H3O+] = 10^-1.987

x = 0.0103 M

at equilibrium [H3O+] = [HCOO-]

so [HCOO-] = 0.0103

at equilibrium concentration of [HCOOH] = 0.55 - x

= 0.55 - 0.0103

= 0.5397 M

now put all these values in above expression

Ka = [0.0103] [0.0103] / 0.5937

= 0.0001 / 0.5397

Ka = 1.96 x 10^-4

queen_honey_blossom answered 1 year ago

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