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Complete this table of values for four aqueous solutions at 25 °C. [H+] [OH-] pH pOH...

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

Solutions

Expert Solution

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

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

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

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

queen_honey_blossom answered 1 year ago

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