Question

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

19 of 20 Part A 0 of each aqueous solution OH stomach acid, 1.7 x 10-13 M Express your answer to two significant figures and include the appropriate units. alue Units Submit Request Answer vPart B urine, 3.8 x 109 M Express your answer to two significant figures and include the appropriate units. Value Units Submit Request Answer Part C orange juice, 2.7 x 1011 M Express your answer to two significant figures and include the appropriate units.

Answer 1

We know,

K_water = [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