A saturated solution of magnesium hydroxide, Mg(OH)_2, is prepared and the exces

Question

A saturated solution of magnesium hydroxide, Mg(OH)_2, is prepared and the excess solid magnesium hydroxide is allowed to settle. A 25.0-mL aliquot of the saturated solution is withdrawn and transferred to an Erlenmeyer flask, and two drops of indicator are added. A 0.00053 M HCl solution (titrant) is dispensed from a buret into the solution (analyte). The solution changes color after the addition of 13.2 mL How many moles of Hydroxide ion are neutralized in the analysis? What is the molar concentration of the hydroxide ion in the saturated solution? What is the molar solubility of magnesium hydroxide? What is the solubility product, K_sp, for magnesium hydroxide? Express the K_sp with the correct number of significant figures?

Explanation / Answer

Answer – We are given, volume of Mg(OH)2 = 25.0 mL , [HCl] = 0.00053 M , volume = 13.2 mL

a) For the calculating neutralized moles of hydroxide ions we need to calculate the moles of HCl.

Reaction – 2 HCl + Mg(OH)2 -----> MgCl­2 + 2 H2O

Moles of HCl = 0.00053 M * 0.0132 L

                      = 6.99*10-6 moles

From the balanced reaction –

2 moles of HCl = 1 moles of Mg(OH)2

So, 6.99*10-6 moles of HCl = ?

= 3.5*10-6 moles of Mg(OH)2

We know,

1 moles of Mg(OH)2 = 2 moles of OH-

So, 3.5*10-6 moles of Mg(OH)2 = ?

= 7.0*10-6 moles of OH-

So, 7.0*10-6 moles of hydroxide ions are neutralized.

b) We calculated moles of hydroxide and we know the volume, so

molarity of OH- = 7.0*10-6 moles /0.025 L

                          = 0.000280 M

The molar concentration of the hydroxide ion is 0.000280 M

c) the molar solubility of the Mg(OH)2 –

we know the molar solubility of the Mg(OH)2 is x

we know the [Mg2+] = 3.5*10-6 moles / 0.025L = 0.000140 M

Mg(OH)2 <---> Mg2+ + 2OH-

                        0.00014 0.000280

So, molar solubility of the Mg(OH)2 = 0.000140 M

d) We know the Ksp expression for the Mg(OH)2

Ksp = [Mg2+][OH-]2

        = 0.000140 *(0.000280)2

        = 1.1*10-11

So, solubility constant for the Mg(OH)2 is 1.1*10-11

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.