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honeytolentino08
06.05.2020 •
Chemistry
If the pOH of a solution is 3.01, what is the concentration of the Mg(OH)2 solution?
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Ответ:
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
Ответ:
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
Ответ: