An alkaline earth hydroxide, M(OH)2, was taken to lab for analysis. The unknown powder was poured into a flask and swirled in room temperature DI water until a saturated solution formed. This solution was then slowly filtered to remove the undissolved solid hydroxide. 28.5 mL of this saturated solution was titrated with 0.173 M HCl (aq). Endpoint required 25.10 mL of the HCl (aq) solution. Calculate the Ksp for this alkaline earth hydroxide.

1.77 * 10^-3

Explanation:

From the titration formula;

Let

CA = concentration of acid

CB = concentration of base

VA = volume of acid

VB = volume of base

NA = number of moles of acid

NB = number of moles of base

The equation of the reaction is;

M(OH)2(aq) + 2HCl(aq) > MCl2(aq) + 2H2O(l)

So;

CAVA/CBVB= NA/NB

CAVANB = CBVBNA

CB= CAVANB/VBNA

CB= 0.173 * 25.10 * 1/28.5 * 2

CB= 4.3423/57

CB= 0.0762 M

This implies that the solubility of M(OH)2 = 0.0762 M

M(OH)2(s) > M^+(aq) + 2OH^-(aq)

So

Ksp = x * (2x)^2

Ksp = 4x^3

x = 0.0762

Ksp= 4(0.0762)^3

Ksp = 1.77 * 10^-3

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