Question

A. What is the pH of an aqueous solution with a hydrogen ion concentration of H 20 x 10-'M? pH = B. What is the hydroxide ion concentration, OH"), in an aqueous solution with a hydrogen ion concentration of (H) = 2.0 x 10-'M? [OH-] = C. A monoprotic weak acid, HA, dissociates in water according to the reaction HA(aq) H(aq) + A+ (aq) 3.00 x 10-M, and The equilibrium concentrations of the reactants and products are (HA) = 0.140 M. H) AT= 3,00 X 10 M. Calculate the K, value for the acid HA.

Answer 1

A)

use:

pH = -log [H+]

= -log (2*10^-7)

= 6.70

Answer: 6.70

B)

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(2*10^-7)

[OH-] = 5.0*10^-8 M

Answer: 5.0*10^-8 M

C)

Ka = [H+][A-]/[HA]

= (3.00*10^-4)*(3.00*10^-4) / (0.140)

= 6.43*10^-7

Answer: 6.43*10^-7

D)

HA dissociates as:

HA -----> H+ + A-

0.14 0 0

0.14-x x x

Ka = [H+][A-]/[HA]

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^-6)*0.14) = 9.466*10^-4

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

so, x = 9.466*10^-4 M

So, [H+] = x = 9.466*10^-4 M

use:

pH = -log [H+]

= -log (9.466*10^-4)

= 3.0238

Answer: 3.02