If the pOH of a solution is 3.01, what is the concentration of the Mg(OH)2 solution?

9.77×10^-4 M

Explanation:

Now you need to recall that pOH=-log [OH^-]

If that be so;

log[OH^-] = - pOH

Then [OH^-] = Antilog (-pOH)

Thus;

[OH^-] = Antilog (-3.01)

[OH^-] = 9.77×10^-4 M

From

Mg(OH)2(s) <> Mg^2+(aq) + 2OH^-(aq)

Let the concentration of the solution= x

Since [Mg^2+] = [2OH^-] = 9.77×10^-4 M = x

Therefore, the concentration of the Mg(OH)2 solution is 9.77×10^-4 M

4.885x10^-4 M

Explanation:

Step 1:

Data obtained from the question.

pOH = 3.01

Step 2:

Determination of the concentration of OH ion [OH-]

This is shown below:

pOH = - Log [OH-]

pOH = 3.01

3.01 = - Log [OH-]

-3.01 = Log [OH-]

Take the anti-log of -3.01

[OH-] = 9.77x10^-4 M

Step 3:

Dissociation equation for Mg(OH)2.

we shall write the equation for the dissociation of Mg(OH)2. This is show below

Mg(OH)2 —> Mg^2+ + 2OH^-

Step 4:

Determination of the concentration of the Mg(OH)2 solution.

From the balanced equation above,

1 mole of Mg(OH)2 dissociate to produce 2 moles of OH-.

Therefore xM of Mg(OH)2 will dissociate to produce 9.77x10^-4 M i.e

xM of Mg(OH)2 = (9.77x10^-4)/2

xM of Mg(OH)2 = 4.885x10^-4 M

Therefore, the concentration of the Mg(OH)2 solution is 4.885x10^-4 M

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