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

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6

i do not get it   at all