Question

What concentration of magnesium hydroxide will make a solution of pH=10.87

Answer 1

....................Mg(OH)2 ==> Mg^2+ + 2OH^-

I....................solid.................0..............0
C...................solid.................x..............2x
E....................solid.................x.............2x
Ksp = 1.8|E-11 = (Mg^2+)(OH^-)^2

You want pH = 10.87 so pOH = 14-10.87 = 3.13
pOH = -log(OH^-) = -log(3.13) so (OH^-) = 7.4E-4 M
Plug in 7.4E-4 for OH^- in the Ksp expression and solve for x = [Mg(OH)2]

Answer 2

Wait I'm sorry what does the e-4 stand for?

Answer 3

Wait nvm that last question

Answer 4

[AI-generated answer — Explain Bot]

To determine the concentration of magnesium hydroxide (Mg(OH)2) required to make a solution with a pH of 10.87, we need to consider the dissociation of magnesium hydroxide in water.

Magnesium hydroxide dissociates into one magnesium ion (Mg2+) and two hydroxide ions (OH-) in water according to the following equation:

Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq)

The equilibrium constant expression for this dissociation is:

Ksp = [Mg2+][OH-]^2

Given that the desired pH is 10.87, we can use the fact that pH is related to the concentration of hydroxide ions (OH-) in solution. In a neutral solution, the concentration of hydroxide ions is equal to the concentration of hydronium ions (H+), which can be calculated using the equation:

pOH = 14 - pH

In this case, pOH = 14 - 10.87 = 3.13

Now, we convert pOH to the concentration of hydroxide ions using the equation:

[OH-] = 10^(-pOH)

[OH-] = 10^(-3.13)

[OH-] ≈ 7.23 x 10^(-4) M

Since the magnesium hydroxide dissociates into two hydroxide ions, the final concentration of magnesium hydroxide will be half the concentration of hydroxide ions:

[Mg(OH)2] ≈ 3.62 x 10^(-4) M

Therefore, a concentration of approximately 3.62 x 10^(-4) M of magnesium hydroxide is required to make a solution with a pH of 10.87.