A)
we have below equation to be used:
[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)/(6.9*10^-9)
[OH-] = 1.449*10^-6 M
we have below equation to be used:
pH = -log [H+]
= -log (6.9*10^-9)
= 8.1612
we have below equation to be used:
pOH = -log [OH-]
= -log (1.449*10^-6)
= 5.8388
Answers:
[H+] = 6.9*10^-9
[OH-] = 1.449*10^-6
pH = 8.1612
pOH = 5.8388
B)
we have below equation to be used:
[H+] = Kw/[OH-]
Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC
[H+] = (1.0*10^-14)/[OH-]
[H+] = (1.0*10^-14)/9.0E-10
[H+] = 1.111*10^-5
we have below equation to be used:
pH = -log [H+]
= -log (1.111*10^-5)
= 4.9542
we have below equation to be used:
pOH = -log [OH-]
= -log (9*10^-10)
= 9.0458
Answers:
[H+] = 1.111*10^-5
[OH-] = 9*10^-10
pH = 4.9542
pOH = 9.0458
C)
POH = 14 - pH
= 14 - 3.92
= 10.08
we have below equation to be used:
pH = -log [H+]
3.92 = -log [H+]
log [H+] = -3.92
[H+] = 10^(-3.92)
[H+] = 1.202*10^-4 M
we have below equation to be used:
pOH = -log [OH-]
10.08 = -log [OH-]
log [OH-] = -10.08
[OH-] = 10^(-10.08)
[OH-] = 8.318*10^-11 M
Answers:
[H+] = 1.202*10^-4
[OH-] = 8.318*10^-11
pH = 3.92
pOH = 10.08
D)
we have below equation to be used:
PH = 14 - pOH
= 14 - 2
= 12
we have below equation to be used:
pH = -log [H+]
12 = -log [H+]
log [H+] = -12
[H+] = 10^(-12)
[H+] = 1*10^-12 M
we have below equation to be used:
pOH = -log [OH-]
2 = -log [OH-]
log [OH-] = -2
[OH-] = 10^(-2)
[OH-] = 1*10^-2 M
Answers:
[H+] = 1.00*10^-12
[OH-] = 1.00*10^-2
pH = 12.00
pOH = 2.00
Feel free to comment below if you have any doubts or if this answer do not work
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