In an aqueous solution of 0.020 M HCl under standard conditions (T=25 C).... (Show & explain work) i) What is the H + concentration? ii) What is the pH of the solution? What about the pOH? iii) What is the OH - of the solution? Solution i) [H+] = [HCl] = 0.020 M Answer: 0.020 M ii) 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^-2) [OH-] = 5*10^-13 M use: pH = -log [H+] = -log (2*10^-2) = 1.699 use: pOH = -log [OH-] = -log (5*10^-13) = 12.301 Answer: pH = 1.70 pOH = 12.30 iii) As calculated above, [OH-] = 5.0*10^-13 M .

1. In an aqueous solution of 0.020 M HCl under standard conditions (T=25 C)....
(Show & explain work)
i) What is the H +
concentration?
ii) What is the pH of the solution? What about the pOH?
iii) What is the OH -
of the solution?
Solution
i)
[H+] = [HCl] = 0.020 M
Answer: 0.020 M
ii)
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^-2)
[OH-] = 5*10^-13 M
use:
pH = -log [H+]

2. = -log (2*10^-2)
= 1.699
use:
pOH = -log [OH-]
= -log (5*10^-13)
= 12.301
Answer:
pH = 1.70
pOH = 12.30
iii)
As calculated above,
[OH-] = 5.0*10^-13 M