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For each strong acid solutions, determine [H3O+],[OH−], and pH. 1) 0.24M HCl 2)2.5x10^-2 M HNO3 3)...

For each strong acid solutions, determine [H3O+],[OH−], and pH.

1) 0.24M HCl

2)2.5x10^-2 M HNO3

3) A solution that is 7.5x10^-2 M in HBr and 2.2x10^-2 M in HNO3

Solutions

Expert Solution

1)

[H+] = [HCl] = 0.24 m

use:

[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)/(0.24)

[OH-] = 4.167*10^-14 M

use:

pH = -log [H+]

= -log (0.24)

= 0.6198

use:

pOH = -log [OH-]

= -log (4.167*10^-14)

= 13.3802

[H3O+] = 0.24

[OH-] = 4.17*10^-14

pH = 0.620

2)

[H+] = [HNO3] = 2.5*10^-2 M

use:

[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)/(2.5*10^-2)

[OH-] = 4*10^-13 M

use:

pH = -log [H+]

= -log (2.5*10^-2)

= 1.6021

use:

pOH = -log [OH-]

= -log (4*10^-13)

= 12.3979

[H3O+] = 2.5*10^-2

[OH-] = 4.0*10^-13

pH = 1.60

3)

[H+] = [HBr] + [HNO3]

= 7.5*10^-2 M + 2.2*10^-2 M

= 9.7*10^-2 M

use:

[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)/(9.7*10^-2)

[OH-] = 1.031*10^-13 M

use:

pH = -log [H+]

= -log (9.7*10^-2)

= 1.0132

use:

pOH = -log [OH-]

= -log (1.031*10^-13)

= 12.9868

[H3O+] = 9.7*10^-2

[OH-] = 1.03*10^-13

pH = 1.013

queen_honey_blossom answered 2 years ago
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