Question

There are four aqueous solutions at 25 degrees celsius.

Solution A

Given [H⁺] = 1.3 x 10^-9 M, find [OH^-], pH, and pOH.

Solution B

Given [OH^-] = 0.098 M, find [H⁺], pH, and pOH.

Solution C

Given pH = 8.16, find [H⁺], [OH^-], and pOH.

Solution D

Given pOH = 1.03, find [H⁺], [OH^-], and pH.

Answer 1

SolutionA:

[H+] = 1.3 x 10^-9M

pH = -logH+

So, pH = -log1.3 x 10^-9

pH = 8.88

pOH = 14 - pH

pOH = 14 - 8.88

pOH = 5.12

[OH-] = 10^-pOH

[OH-] - 10^-5.12

[OH-] = 7.5 x 10^-6M

Solution B:

[OH-] = 0.098

pOH = -log0.098

pOH = 1.01

pH = 14 - pOH

pH = 14 - 1.01

pH = 12.99

[H+] = 10^-pH

[H+] = 10^-12.99

[H+] = 1.02 x 10^-13M

Solution C:

pH = 8.16

[H+] = 10^-8.16 as [H+] = 10^-pH

[H+] = 6.92 x 10^-9M

pOH = 14 - 8.16

pOH = 5.84

[OH-] = 10^-pOH

[OH-] = 10^-5.84

[OH-] = 1.4 x 10^-6M

Solution D:

pOH = 1.03

[OH-] = 10^-1.03

[OH-] = 0.093

pH = 14 - 1.03

pH = 12.97

[H+] = 10^-12.97

[H+] = 1.07 x 10^-13M