Question

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a. For the strong acid solution 5.7×10−3 M HNO3, determine [H3O+] and [OH−]. then find pH...

a. For the strong acid solution 5.7×10−3 M HNO3, determine [H3O+] and [OH−]. then find pH

b.For the strong acid solution 0.0826 M HBr, determine [H3O+] and [OH−]. then find pH

c. For the strong acid solution 9.9×10−4 M HI, determine [H3O+] and [OH−]. Then find pH.

d. For the strong acid solution 0.0650 M HCl, determine [H3O+] and [OH−]. Then find pH.

Solutions

Expert Solution

a)

[H3O+] = [H+] = 5.7*10^-3 M

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)/(5.7*10^-3)

[OH-] = 1.754*10^-12 M

we have below equation to be used:

pH = -log [H+]

= -log (5.7*10^-3)

= 2.2441

[H3O+] = 5.7*10^-3

[OH-] = 1.754*10^-12

pH = 2.2441

b)

[H3O+] = [H+] = 0.0826 M

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)/(8.26*10^-2)

[OH-] = 1.211*10^-13 M

we have below equation to be used:

pH = -log [H+]

= -log (8.26*10^-2)

= 1.083

[H3O+] = 8.26*10^-2

pH = 1.083

pOH = 12.917

c)

[H+]=[H3O+] = 9.9*10^-4 M

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)/(9.9*10^-4)

[OH-] = 1.01*10^-11 M

we have below equation to be used:

pH = -log [H+]

= -log (9.9*10^-4)

= 3.0044

[H+] = 9.9*10^-4

[OH-] = 1.01*10^-11

pH = 3.0044

d)

[H3O+] =[H+] = 0.0650

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)/(6.5*10^-2)

[OH-] = 1.538*10^-13 M

we have below equation to be used:

pH = -log [H+]

= -log (6.5*10^-2)

= 1.1871

[H3o+] = 6.5*10^-2

[OH-] = 1.538*10^-13

pH = 1.1871

queen_honey_blossom answered 2 years ago
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