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

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Chemistry

In the laboratory, a general chemistry student measured the pHof a 0.550 M aqueous solution of formicacid, HCOOH to be1.987.

Use the information she obtained to determine the K_a forthis acid.

K_a(experiment) =

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4.4 Ratings (730 Votes)

we have to use ICE table to calculate this

HCOOH + H2O <-------> HCOO^- + H3O⁺

dissociation constant Ka expression for this equation

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

Â Â Â ICE Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ HCOOH ] Â Â [ HCOO^- ] Â Â Â Â Â Â Â Â [ H3O⁺ ]

initial concentrationÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0.55Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Change in concentrationÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -x

equilibrium concentrationÂ Â Â Â Â Â Â Â Â Â 0.55 - xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

x = [H3O⁺]

pH = -log[H3O+]Â Â Â Â Â

pH he has given 1.987

1.987 = -log[H3O+]Â Â Â Â

[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

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