Question

When a metallic hydroxide is dissolved in water the hydroxide ion concentration depends upon the formula of the metallic hydroxide. When completely dissociated, the hydroxide ion concentration will be the concentration of the whole compound times the number of hydroxide ions units in the formula. 1. Calculate the concentration of hydroxide ions and hydronium ions in the following basic solutions: (a) A solution prepared by dissolving 200 g of magnesium hydroxide in enough water to make 4.00 L of solution. (b) A solution containing 2.00 mol of strontium hydroxide is diluted up to 3.00 L (c) A solution prepared by dissolving 150 g of aluminum hydroxide in enough water to prepare .00 L of solution (assumer the aluminum hydroxide is soluble)

Answer 1

(a)

Mass of magnesium hydroxide = 200 g.

Molar mass of magnesium hydroxide = 58.3 g/mol

Moles of magnesium hydroxide = mass / molar mass = 200 / 58.3 = 3.43 mol

Volume of solution = 4.00 L

Molarity = moles of solute / volume fo slution in L

M = 3.43 / 4.00

M = 0.858 M

Mg(OH)2 (aq) ---------------> Mg^2+ (aq) + 2 OH^- (aq)
From the above equation,

[OH^-] = 2 x [Mg(OH)2] = 2 x 0.858 = 1.72 M

For aqueous solutions, at 298 K,

[H3O^+] [OH-] = 10^-14

[H3O^] = 10^-14 / [OH-] = 10^-14 / 1.72 = 5.83 x 10^-15 M

Therefore,

[OH-] = 1.72 M

[H3O^+] = 5.83 x 10^-15 M

(b)

Molarity = moles / volume in L = 2.00 / 3.00 = 0.667 M

Sr(OH)2 (aq) ---------------> Sr^2+ (aq) + 2 OH^- (aq)

From the above equation,

[OH^-] = 2 x [Sr(OH)2] = 2 x 0.667 = 1.33 M

[H3O^+] = 10^-14 / [OH-] = 10^-14 / 1.33 = 7.50 x 10^-15 M

Therefore,

[OH^-] = 1.33 M

[H3O^+] = 7.50 x 10^-15 M

(c)

Mass of Al(OH)3 = 150 g.

Molar mass of Al(OH)3 = 78 g/mol

Moles of Al(OH)3 = mass / molar mass = 150. / 78 = 1.92 mol

Volume of Solution = 1.00 L ( assumed 1.00 L as it is not printed correctly in the question )

Molarity = moles / volume in L = 1.92 / 1.00

M = 1.92 M

Al(OH)3 (aq) --------------> Al^3+ (aq) + 3 OH^- (aq)

From the above equation,

[OH^-] = 3 x [Al(OH)3] = 3 x 1.92 = 5.76 M

[H3O^+] = 10^-14 / 5.76 = 1.74 x 10^-15 M

Therefore,

[OH-] = 5.76 M

[H3O^+] = 1.74 x 10^-15 M