Question

i need help with both :(

33. What is the pH of a 0.75 M Benzoic Acid (HC H3O2) solution? Ka for Benzoic Acid is 6.4 x 10-5 34. What is the pH of a 0.040 M Pyridine (CsH5N) solution? Kb for Pyridine is 1.7 x 10-9

Answer 1

33)

C6H5COOH dissociates as:

C6H5COOH -----> H+ + C6H5COO-

0.75 0 0

0.75-x x x

Ka = [H+][C6H5COO-]/[C6H5COOH]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((6.4*10^-5)*0.75) = 6.928*10^-3

since c is much greater than x, our assumption is correct

so, x = 6.928*10^-3 M

So, [H+] = x = 6.928*10^-3 M

use:

pH = -log [H+]

= -log (6.928*10^-3)

= 2.1594

Answer: 2.16

34)

C5H5N dissociates as:

C5H5N +H2O -----> C5H5NH+ + OH-

4*10^-2 0 0

4*10^-2-x x x

Kb = [C5H5NH+][OH-]/[C5H5N]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.7*10^-9)*4*10^-2) = 8.246*10^-6

since c is much greater than x, our assumption is correct

so, x = 8.246*10^-6 M

So, [OH-] = x = 8.246*10^-6 M

use:

pOH = -log [OH-]

= -log (8.246*10^-6)

= 5.0837

use:

PH = 14 - pOH

= 14 - 5.0837

= 8.9163

Answer: 8.92