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In: Chemistry

For each strong base solution, determine [OH−], [H3O+], pH, and pOH. Part A 6.77×10−3 M LiOH,...

For each strong base solution, determine [OH−], [H3O+], pH, and pOH.

Part A

6.77×10−3 M LiOH, determine [OH−]and [H3O+].

Express your answers using three significant figures. Enter your answers numerically separated by a comma.
[OH−], [H3O+]= M

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Part B

For this solution determine pH and pOH.

Express your answers using three decimal places. Enter your answers numerically separated by a comma.
pH, pOH=

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Part C

0.0312 M Ba(OH)2, determine [OH−]and [H3O+].

Express your answers using three significant figures. Enter your answers numerically separated by a comma.
[OH−], [H3O+]= M

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Part D

For this solution determine pH and pOH.

Express your answers using three decimal places. Enter your answers numerically separated by a comma.
pH, pOH=

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Part E

5.9×10−4 M KOH, determine [OH−]and [H3O+].

Express your answers using two significant figures. Enter your answers numerically separated by a comma.
[OH−], [H3O+]= M

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Part F

For this solution determine pH and pOH.

Express your answers using two decimal places. Enter your answers numerically separated by a comma.
pH, pOH=

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Part G

4.0×10−4 M Ca(OH)2, determine [OH−] and [H3O+].

Express your answers using two significant figures. Enter your answers numerically separated by a comma.
[OH−], [H3O+]= M

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Part H

For this solution determine pH and pOH.

Express your answers using two decimal places. Enter your answers numerically separated by a comma.
pH, pOH=

Solutions

Expert Solution

A)

[OH-] = 6.77*10^-3 M

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)/6.77*10^-3 M

[H+] = 1.48*10^-12 M

6.77*10^-3 , 1.48*10^-12 M

B)

we have below equation to be used:

pH = -log [H+]

= -log (1.477*10^-12)

= 11.831

we have below equation to be used:

pOH = -log [OH-]

= -log (6.77*10^-3)

= 2.169

11.831, 2.169

C)

[OH-] =2*[Ba(OH)2] = 2*0.0312 M = 0.0624 M

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)/0.0624

[H+] = 1.603*10^-13 M

Answer: 0.0624, 1.60*10^-13 M

D)

we have below equation to be used:

pH = -log [H+]

= -log (1.603*10^-13)

= 12.795

we have below equation to be used:

pOH = -log [OH-]

= -log (6.24*10^-2)

= 1.205

12.795, 1.205

only 4 parts at a time

queen_honey_blossom answered 2 years ago
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