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Calculate the [OH−] of each aqueous solution with the following [H3O+]: stomach acid, 2.0×10−2M Express your...

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

stomach acid, 2.0×10−2M Express your answer using two significant figures.

urine, 7.0×10−6M Express your answer using two significant figures.

orange juice, 2.2×10−4M Express your answer using two significant figures.

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

Solutions

Expert Solution

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)/(2*10^-2)

[OH-] = 5.0*10^-13 M

Answer: 5.0*10^-13 M

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)/(7*10^-6)

[OH-] = 1.429*10^-9 M

Answer: 1.4*10^-9 M

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)/(2.2*10^-4)

[OH-] = 4.545*10^-11 M

Answer: 4.5*10^-11 M

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 1 year ago

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