In: Chemistry
Complete this table of values for four aqueous solutions at 25 °C.
[H+] [OH-] pH pOH
solutionA : 8.7x10^-7 M
solutionB: 6.9x10^-9 M
solutionC: 9.75
solution D: 6.39
A)
we have below equation to be used
[OH-] = (1.0*10^-14)/[H3O+]
[OH-] = (1.0*10^-14)/(8.7*10^-7)
[OH-] = 1.149*10^-8 M
we have below equation to be used
pH = -log [H+]
= -log (8.7*10^-7)
= 6.0605
we have below equation to be used
pOH = -log [OH-]
= -log (1.149*10^-8)
= 7.9395
Since pH < pOH, this is acidic in nature
[H+] = 8.7*10^-7
[OH-] = 1.1*10^-8
pH = 6.06
pOH = 7.94
B)
we have below equation to be used
[H+] = (1.0*10^-14)/[OH-]
[H+] = (1.0*10^-14)/6.9E-9
[H+] = 1.449*10^-6
we have below equation to be used
pH = -log [H+]
= -log (1.449*10^-6)
= 5.8388
we have below equation to be used
pOH = -log [OH-]
= -log (6.9*10^-9)
= 8.1612
Since pH < pOH, this is acidic in nature
[H+] = 1.5*10^-6
[OH-] = 6.9*10^-9
pH = 5.84
pOH = 8.16
C)
POH = 14 - pH
= 14 - 9.75
= 4.25
we have below equation to be used
pH = -log [H3O+]
9.75 = -log [H3O+]
[H3O+] = 1.778*10^-10 M
we have below equation to be used
pOH = -log [OH-]
4.25 = -log [OH-]
[OH-] = 5.623*10^-5 M
Since pH > pOH, this is basic in nature
[H+] = 1.8*10^-10
[OH-] = 5.6*10^-5
pH = 9.75
pOH = 4.25
D)
we have below equation to be used
PH = 14 - pOH
= 14 - 6.39
= 7.61
we have below equation to be used
pH = -log [H3O+]
7.61 = -log [H3O+]
[H3O+] = 2.455*10^-8 M
we have below equation to be used
pOH = -log [OH-]
6.39 = -log [OH-]
[OH-] = 4.074*10^-7 M
Since pH > pOH, this is basic in nature
[H+] = 2.5*10^-8
[OH-] = 4.1*10^-7
pH = 7.61
pOH = 6.39