We know,
Kwater = [H3O+] * [OH-]
=> [H3O+] * [OH-] = 1*10-14
=> [H3O+] = 1*10-14 / [OH-]
Part A:
Stomach acid - [OH-] = 1.7*10-13M
So, [H3O+] = 1*10-14 / [OH-] = 1*10-14 / 1.7*10-13 = 0.0588M
Therefore [H3O+] of stomach acid = 0.059M
Part B:
Urine - [OH-] = 3.8*10-9M
So, [H3O+] = 1*10-14 / [OH-] = 1*10-14 / 3.8*10-9 = 2.631*10-6M
Therefore [H3O+] of urine =2.6*10-6M
Part C:
Orange Juice - [OH-] = 2.7*10-11M
So, [H3O+] = 1*10-14 / [OH-] = 1*10-14 / 2.7*10-11 = 3.7*10-4M
Therefore [H3O+] of Orange juice =3.7*10-4M
Part D:
Bile - [OH-] = 3.9*10-6M
So, [H3O+] = 1*10-14 / [OH-] = 1*10-14 / 3.9*10-6 = 2.56*10-9M
Therefore [H3O+] of Bile =2.6*10-9M
Calculate the [H3O+] of each aqueous solution with the following [OH-]: 19 of 20 Part A...
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