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Calculate the [OH−] of each aqueous solution with the following [H3O+]: 1)stomach acid, 5.0×10−2M. Express your...

Calculate the [OH−] of each aqueous solution with the following [H3O+]:

1)stomach acid, 5.0×10−2M. Express your answer using two significant figures.

2)urine, 3.0×10−6M. Express your answer using two significant figures.

3)orange juice, 1.6×10−4M. Express your answer using two significant figures.

4)bile, 6.0×10−9M.Express your answer using two significant figures.

Solutions

Expert Solution

1)

we have below equation to be used:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(5*10^-2)

[OH-] = 2.0*10^-13 M

Answer: 2.0*10^-13 M

2)

we have below equation to be used:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(3*10^-6)

[OH-] = 3.3*10^-9 M

Answer: 3.3*10^-9 M

3)

we have below equation to be used:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(1.6*10^-4)

[OH-] = 6.2*10^-11 M

Answer: 6.2*10^-11 M

4)

we have below equation to be used:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(6*10^-9)

[OH-] = 1.7*10^-6 M

Answer: 1.7*10^-6 M

queen_honey_blossom answered 2 years ago
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