Experiment 22 (Part A) Molar Solubility and Solubility Product of Calcium Hydrox

Question

Experiment 22

(Part A) Molar Solubility and Solubility Product of Calcium Hydroxide and (Part B) Molar Solubility of Calcium Hydroxide in the Presence of a Common Ion.

These are the calculations my lab partners and I came up with, and I was just hoping to see if they were correct and if a tutor here could help write out and explain the process of how the calculations are done so I can include them in my lab report? My teacher and I both like to do things step-by-step, so an explanation like that would be very helpful!
I will also include a blank report sheet of each so that you can see what the values are representing. If you could also offer your thoughts for accounting the difference between A 11 and B 10 that would be great so I can compare my answer. Thank you!

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Explanation / Answer

Common ion effect

When a common ion is added to a solution of a sparingly soluble salt. Its solubility decreases.

In Ist part,

moles of HCl used = moles of OH- present in solution

moles = molarity (M) x volume (L)

for Ca(OH)2

in solution we have for every mole of Ca2+, 2 moles of OH-

so, moles of Ca2+ = 1/2 moles of OH-

molar solubility of Ca(OH)2 = moles Ca2+/volume of initial solution taken

The calculations shown above are correct.

In IInd part, effect of common ion

when Ca(OH)2 in presence of CaCl2 was studied.

molar solubility was reduced as per the expectation

Excess Ca2+ in solution,

Ca(OH)2(s) <==> Ca2+(aq) + 2OH-(aq)

CaCl2(aq) --> Ca2+(aq) + 2Cl-(aq)

on the right handside, causes the Ca(OH)2 reaction to shift backwards and form more undissociated Ca(OH)2. this reduces the solubility of Ca(OH)2 in solution.

Calculations are again correct.

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