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

it is the most commonly accepted one

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.