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
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