problem 11.39 chapter 11 post Calculate the [H3O+] of each solution with the fol

Question

problem 11.39 chapter 11 post
Calculate the [H3O+] of each solution with the following [OH-]
A) coffee 1.0 ×10^-9 M
B) soap 1.5×10^-6 M
C) cleanser 2.5×10^-5 M
D) lemon juice 3.0×10^-13 M

problem 11.40
calculate the [H3O+] of each aqueous solution with the following [OH-]
A) NaOH 1.4×10^-2 M
B) milk of magnesia 1.4×10^-5 M
C) aspirin 2.6×10^-11 M
D) seawater 4.0×10^-6 M

Calculate the [OH-] of each aqueous solution with following [H3O+]
A) stomach acid 5.0×10^-2 M
B) urine 6.0×10^-6 M
C) orange juice 1.4×10^-4 M
D) bile 8.5×10^-9 M

calculate the [OH-] of each aqueous solution with the following [H3O+]
A) baking soda 1.0×10^-8 M
B) blood 4.5×10^-8 M
C) milk 5.0×10^-7 M
D) pancreatic juice 3.9×10^-9 M

problem 11.107
what are the [H3O+] and [OH-] for a solution with each of the following pH values
A) pH=3.2
B) ph = 6.0
C) pH= 8.72
D) pH= 10.00

Explanation / Answer

I will do one of these

Calculate the [H3O+] of each solution with the following [OH-]
A) coffee 1.0 ×10^-9 M
B) soap 1.5×10^-6 M
C) cleanser 2.5×10^-5 M
D) lemon juice 3.0×10^-13 M

When the [OH-] is given we can calculate pOH as

pOH = -log [OH-]

a) pOH = - log 1.0 ×10-9

pOH = 9

pH = 14 - pOH

pH = 14 - 9

pH = 5

[H3O+] = 10-pH

[H3O+] = 10-5

[H3O+] = 1 x 10-5 M

b)

a) pOH = - log 1.5 ×10-6

pOH = 5.82

pH = 14 - pOH

pH = 14 - 5.82

pH = 8.17

[H3O+] = 10-pH

[H3O+] = 10-8.17

[H3O+] = 6.67 x 10-9 M

c)

a) pOH = - log 2.5 ×10-5

pOH = 4.6

pH = 14 - pOH

pH = 14 - 4.6

pH = 9.38

[H3O+] = 10-pH

[H3O+] = 10-9.38

[H3O+] = 4 x 10-10 M

d)

a) pOH = - log 3 ×10-13

pOH = 12.52

pH = 14 - pOH

pH = 14 - 12.52

pH = 1.478

[H3O+] = 10-pH

[H3O+] = 10-1.478

[H3O+] = 3.33 x 10-2 M

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