For each of the following strong base solutions,determine [OH], [H3O+], pH, pOH: 0.18M NaOH 1.7*10^-3m Ca(OH)2...

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

0.18M NaOH

1.7*10^-3m Ca(OH)2

5.0*10^-4M Sr(OH)2

8.8*10^-5M KOH

Solutions

Expert Solution

1)
[OH-] = [NaOH] = 0.18 M

use:
[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)/0.18
[H+] = 5.556*10^-14 M

use:
pH = -log [H+]
= -log (5.556*10^-14)
= 13.2553

use:
pOH = -log [OH-]
= -log (0.18)
= 0.7447

Answers:
[H+] = 5.6*10^-14
[OH-] = 0.18
pH = 13.26
pOH = 0.745

2)
[OH-] = 2*[Ca(OH)2]
= 2*1.7*10^-3 M
= 3.4*10^-3 M

use:
[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)/3.4*10^-3
[H+] = 2.941*10^-12 M

use:
pH = -log [H+]
= -log (2.941*10^-12)
= 11.5315

use:
pOH = -log [OH-]
= -log (3.4*10^-3)
= 2.4685

Answers:
[H+] = 2.9*10^-12
[OH-] = 3.4*10^-3
pH = 11.53
pOH = 2.47

3)
[OH-] = 2*[Sr(OH)2]
= 2*5.0*10^-4 M
= 1.0*10^-3 M

use:
[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)/1*10^-3
[H+] = 1*10^-11 M

use:
pH = -log [H+]
= -log (1*10^-11)
= 11

use:
pOH = -log [OH-]
= -log (1*10^-3)
= 3

Answers:
[H+] = 1.0*10^-11
[OH-] = 1.0*10^-3
pH = 11.00
pOH = 3.00

4)
[OH-] = [KOH] = 8.8*10^-5 M

use:
[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)/8.8*10^-5
[H+] = 1.136*10^-10 M

use:
pH = -log [H+]
= -log (1.136*10^-10)
= 9.9445

use:
pOH = -log [OH-]
= -log (8.8*10^-5)
= 4.0555

Answers:
[H+] = 1.1*10^-10
[OH-] = 8.8*10^-5
pH = 9.94
pOH = 4.06

queen_honey_blossom answered 5 months ago

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