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Bookmarks Develop Window Help 10 saplinglearning.com University of California - Merced- CHEM 10 (INCLUSIVE ACCESS)-Fall17-ROBLEDO: Focus 6: Topics 6A-6F Jump to... 30/2017 12:00 AM 0 17.1/100 o 10/29/2017 06:26 PM Gradeboo Print CalculatorPeriodic Table ion 6 of 26 Sapling Learning Map Complete this table of values for four aqueous solutions at 25 . OH pH pOH Given: Solution A: 9.1x 10-7 Given Number Sclution B 0.000048 Number Number Solution C: 7.93 Given: Solution D: 8.07 O Previous Give Up & View Sounon Check Answer 0 Next Exit about us careers privacy policy terms of use contact us help

Explanation / Answer

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)/(9.1*10^-7)

[OH-] = 1.099*10^-8 M

we have below equation to be used:

pH = -log [H+]

= -log (9.1*10^-7)

= 6.041

we have below equation to be used:

pOH = -log [OH-]

= -log (1.099*10^-8)

= 7.959

Answers:

[H+] = 9.1*10^-7

[OH-] = 1.099*10^-8

pH = 6.041

pOH = 7.959

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)/4.8E-5

[H+] = 2.083*10^-10

we have below equation to be used:

pH = -log [H+]

= -log (2.083*10^-10)

= 9.6812

we have below equation to be used:

pOH = -log [OH-]

= -log (4.8*10^-5)

= 4.3188

Answers:

[H+] = 2.083*10^-10

[OH-] = 4.8*10^-5

pH = 9.6812

pOH = 4.3188

C)

POH = 14 - pH

= 14 - 7.93

= 6.07

we have below equation to be used:

pH = -log [H+]

7.93 = -log [H+]

log [H+] = -7.93

[H+] = 10^(-7.93)

[H+] = 1.175*10^-8 M

we have below equation to be used:

pOH = -log [OH-]

6.07 = -log [OH-]

log [OH-] = -6.07

[OH-] = 10^(-6.07)

[OH-] = 8.511*10^-7 M

Answers:

[H+] = 1.175*10^-8

[OH-] = 8.511*10^-7

pH = 7.93

pOH = 6.07

D)

we have below equation to be used:

PH = 14 - pOH

= 14 - 8.07

= 5.93

we have below equation to be used:

pH = -log [H+]

5.93 = -log [H+]

log [H+] = -5.93

[H+] = 10^(-5.93)

[H+] = 1.175*10^-6 M

we have below equation to be used:

pOH = -log [OH-]

8.07 = -log [OH-]

log [OH-] = -8.07

[OH-] = 10^(-8.07)

[OH-] = 8.511*10^-9 M

Answers:

[H+] = 1.175*10^-6

[OH-] = 8.511*10^-9

pH = 5.93

pOH = 8.07

Feel free to comment below if you have any doubts or if this answer do not work

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