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For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH. 1. 6.77×10−3 M...

For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH.

1. 6.77×10−3 M LiOH,

2. 0.0412 M Ba(OH)2

3. 5.9×10−4 M KOH

4. 3.0×10−4 M Ca(OH)2

Solutions

Expert Solution

1)

[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)/0.00677

[H+] = 1.477*10^-12 M

we have below equation to be used:

pH = -log [H+]

= -log (1.477*10^-12)

= 11.83

we have below equation to be used:

pOH = -log [OH-]

= -log (6.77*10^-3)

= 2.17

[H3O+] = 1.477*10^-12 M

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

pH = 11.83

pOH = 2.17

2)

[OH-] = 2*[Ba(OH)2] = 2*0.0412 M = 8.24*10^-2

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)/0.0824

[H+] = 1.214*10^-13 M

we have below equation to be used:

pH = -log [H+]

= -log (1.214*10^-13)

= 12.9159

we have below equation to be used:

pOH = -log [OH-]

= -log (8.24*10^-2)

= 1.0841

[H3O+] = 1.214*10^-13

[OH-] = 8.24*10^-2

pH = 12.92

pOH = 1.08

3)

[OH-] = 5.9*10^-4

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)/5.9*10^-4

[H+] = 1.695*10^-11 M

we have below equation to be used:

pH = -log [H+]

= -log (1.695*10^-11)

= 10.7709

we have below equation to be used:

pOH = -log [OH-]

= -log (5.9*10^-4)

= 3.2291

[H3O+] = 1.695*10^-11

[OH-] = 5.9*10^-4

pH = 10.77

pOH = 3.23

4)

[OH-] = 2*[Ca(OH)2]= 2* 3.0*10^-4 = 6*10^-4

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)/6.0*10^-4

[H+] = 1.667*10^-11 M

we have below equation to be used:

pH = -log [H+]

= -log (1.667*10^-11)

= 10.7782

we have below equation to be used:

pOH = -log [OH-]

= -log (6*10^-4)

= 3.2218

[H3O+] = 1.667*10^-11

[OH-] = 6*10^-4

pH = 10.78

pOH = 3.22

queen_honey_blossom answered 2 years ago
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