calculate either [H3O+] or [OH-] for the solutions below.

Question

1)[OH-] = 1.77 x 10^-7 M

[H3O+] = ?M

2)[H3O+] = 8.25 x 10^-9

[OH-] = ?M

3)[H3O+] = 0.000653 M

[OH-] = ?M

Answer 1

K_w = 1x10^-14 = [H3O⁺][OH^-]

[OH^-] = 1.77x10^-7

[H3O+] = 1x10¹⁴/1.77x10^-7 = 5.65x10^-8 M

 

Do the others the same way.  

Answer 2

(1)

 

[OH^-] = 1.77 x 10^-7 

 

pOH = -log[OH^-] = -log(1.77 x 10^-7) = 6.752

 

pH = 14 - pOH = 14 - 6.752 = 7.248

 

[H₃O⁺] = 10^-pH = 10^-7.248 = 5.649 x 10^-8

 

(2)

 

[H₃O⁺] = 8.25 x 10^-9

 

pH = -log(8.25 x 10^-9) = 8.084

 

pOH = 14 - pH = 14 - 8.084 = 5.916

 

[OH^-] = 10^-5.916 = 1.213 x 10^-6

 

(3)

 

Solve the same way as (2).  You should be able to finish from here.