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Mg(OH)_2 is a slightly soluble salt. In order to determine the of Mg(OH)_2, a ti

ID: 973040 • Letter: M

Question

Mg(OH)_2 is a slightly soluble salt. In order to determine the of Mg(OH)_2, a titration experiment was performed. A 25.00 mL sample of a saturated Mg(OH)_2 solution was titrated to the endpoint with 3.11 mL of a standardized 0.0025 M HCI solution. The molar mass of Mg(OH)_2 is 58.3 g/mol. a) Write the equilibrium reaction for solid Mg(OH)_2 dissolving in water. Indicate a phase for each species in your reaction. b) Calculate the moles of OH^- in the solution. c) Calculate the moles of Mg^2+ in the solution. d) Calculate K_sp for Mg(OH)_2.

Explanation / Answer

a)

b) Calculation for the moles of OH- ions:

it is given that

volume (V1) of Mg (OH)2 = 25 mL

volume (V2)of HCl consumed = 3.11 mL

molarity of HCl (M2)= 0.0025M

Therefore M1V1 = M2V2

Implies M1×25 = 0.0025×3.11

M1 = (0.0025×3.11)/25

    = 0.0003 M

Therefore no of moles of OH = 2× molarity ( as per the stiochiometry)

                                                       = 2×0.0003

     = 0.0006 moles

c) Calculation for the moles of Mg+2 ions:

moles of Mg+2 ions = 1× molarity (as per the stichiometry)

= 0.0003 moles

d) Calculation Ksp:

Ksp = [ Mg+2] 2× [OH-]

    = 0.0003× 2×0.0006

= 0.00000036