We investigated solubility of Ca(OH)_2 in water by pH measurement. Now if we hav

Question

We investigated solubility of Ca(OH)_2 in water by pH measurement. Now if we have a corrected pH = 12.2 for a saturated Ca(OH)_2 solution at 25 degree C What is the OH^- concentration in the solution? _______ What is the solubility constant of Ca(OH)_2 calculated from the data above? ________K_sp = ______ According to your measurement, when heating the saturated Ca(OH)_2 solution from room temperature (T_1) to 50 degree C, the solution pH is _______. It indicates that the water solubility of Ca(OH)_2 at higher temperature is ____________. increased decreased unchanged The result also tells us that the dissolution of Ca(OH)_2 in water ________ requires absorption of heat, Delta H > 0 will release heat, Delta H

Explanation / Answer

Q1.

pH = 12.2

i)

OH-

pOH = 14-pH = 14-12.2 = 1.8

[OH-] = 10^-pOH = 10^-1.8 = 0.01584 M

ii=

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

[Ca+2] = 1/2[OH-]

so

Ksp = 1/2*[OH-][OH-]^2

Ksp = 0.5*([OH-])^3 = 0.5*(0.01584^3)

Ksp = 0.00000198717

iii)

pH increases

therefore, OH- increases

meaning that Ca(OH)2 solubility will decrease as T increases

then

dissolutino od Ca(OH)2 does NOT requiers heat, i.e. i is releasing energy

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