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Calculating [H.]

DETERMINE THE CONCENTRATION OF HOF EACH SOLUTION
» [OH-] = 1 x 10-IM
CH'] =
» pH = 2
[H] =
» [OH ) = 1 x 10-8 M
(H"]=
» POH= 9
[H] -

Calculating [H.] DETERMINE THE CONCENTRATION OF HOF EACH SOLUTION » [OH-] = 1 x 10-IM-example-1
User NitinSingh
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1 Answer

1 vote

Answer:

The answer to your question is given below below:

Step-by-step explanation:

1. Data obtained from the question:

Concentration of Hydroxide ion, [OH-] = 1x10^-1 M

Concentration of Hydrogen ion, [H+] =..?

The concentration of Hydrogen ion and hydroxide ion are related with the following formula:

[H+] x [OH-] = 1x10^-14

[H+] x 1x10^-1 = 1x10^-14

Divide both side by 1x10^-1

[H+] = 1x10^-14 / 1x10^-1

[H+] = 1x10^-13M

2. Data obtained from the question include:

pH = 2

Hydrogen ion concentration, [H+] =..?

pH = - log [H+]

2 = - log [H+]

-2 = log [H+]

[H+] = antilog (-2)

[H+] = 0.01M

3. Data obtained from the question:

Concentration of Hydroxide ion, [OH-] = 1x10^-8 M

Concentration of Hydrogen ion, [H+] =..?

The concentration of Hydrogen ion and hydroxide ion are related with the following formula:

[H+] x [OH-] = 1x10^-14

[H+] x 1x10^-8 = 1x10^-14

Divide both side by 1x10^-8

[H+] = 1x10^-14 / 1x10^-8

[H+] = 1x10^-6M

4. Data obtained from the question include:

pOH = 9

Hydrogen ion concentration, [H+] =..?

First we shall determine the pH.

The pH and pOH are related with the following formula:

pH + pOH = 14

pH + 9 = 14

Collect like terms

pH = 14 - 9

pH = 5

Now, we can obtain the [H+] as follow:

pH = - log [H+]

pH = 5

5 = - log [H+]

-5 = log [H+]

[H+] = antilog (-5)

[H+] = 1x10^-5M

User Marna
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