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Liula [17]
3 years ago
9

A solution has a oh- = 1 x 10-5 m what are the h30 and the ph of the solution

Chemistry
2 answers:
ruslelena [56]3 years ago
6 0

Answer : The H_3O^+ concentration and the pH of the solution is, 1\times 10^{-9}M and 9 respectively.

Explanation:

First we have to calculate the pOH.

pOH=-\log [OH^-]

pOH=-\log (1\times 10^{-5})

pOH=5

Now we have to calculate the pH.

pH+pOH=14\\\\pH=14-pOH\\\\pH=14-5=9

Now we have to calculate the H_3O^+ concentration.

pH=-\log [H_3O^+]

9=-\log [H_3O^+]

[H_3O^+]=1\times 10^{-9}M

Therefore, the H_3O^+ concentration and the pH of the solution is, 1\times 10^{-9}M and 9 respectively.

Deffense [45]3 years ago
3 0

I think it's easiest to find the pOH from the given [OH-] first.

-log(1x10^-5)

pOH=5

Then find the pH.

pOH+pH=14

5+pH=14

pH=9

Then find the [H+] using the pH.

antilog(-9) (if you dont have an antilog button use 10^-9)

[H+]=1x10^-9

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Explanation:

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8 0
3 years ago
Carbon monoxide (CO) reacts with hydrogen (H2) to form methane (CH4) and water (H2O).
Artemon [7]

Answer:

5.9x10^-2 M

Explanation:

Step 1:

Data obtained from the question. This includes the following:

Concentration of CO, [CO] = 0.30 M

Concentration of H2, [H2] = 0.10 M

Concentration of H2O, [H2O] = 0.020 M

Equilibrium constant, K = 3.90

Concentration of CH4, [CH4] =..?

Step 2:

The balanced equation for the reaction. This is given below:

CO(g) + 3H2(g) <=> CH4(g) + H2O(g)

Step 3:

Determination of the concentration of CH4.

The expression for equilibrium constant of the above equation is given below:

K = [CH4] [H2O] / [CO] [H2]^3

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Cross multiply to express in linear form

[CH4] x 0.02= 3.9 x 0.3 x (0.1)^3

Divide both side by 0.02

[CH4] = 3.9 x 0.3 x (0.1)^3 /0.02

[CH4] = 5.9x10^-2 M

Therefore, the equilibrium concentration of CH4 is 5.9x10^-2 M

5 0
3 years ago
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