The pH of a 1.4 M solution of acrylic acid (HC3H,CO2) is measured to be 2.05...

Question

Question

science chemistry

Answer 1

Answer #1

use:

pH = -log [H+]

2.05 = -log [H+]

[H+] = 8.913*10^-3 M

HC3H3CO2 dissociates as:

HC3H3CO2 -----> H+ + C3H3CO2-

1.4 0 0

1.4-x x x

Ka = [H+][C3H3CO2-]/[HC3H3CO2]

Ka = x*x/(c-x)

Ka = 8.913*10^-3*8.913*10^-3/(1.4-8.913*10^-3)

Ka = 5.71*10^-5

Answer: 5.7*10^-5

Answer 2

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