13. A sample of solid Ca(OH) 2 was stirred in water at a certain temperature unt

Question

13. A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)2 as it could hold. A90.8-mL sample of this solution was withdrawn and titrated with 0.0558 M HBr. It required 61.7 mL of the acid solution for neutralization.

(a) What was the molarity of the Ca(OH)2 solution?

____ M



(b) What is the solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100 mL of solution?

____ g/100mL

15. Indicate the concentration of each ion present in the solution formed by mixing the following. (Assume that the volumes are additive.)

(a) 10 mL of 0.100 M HCl and 10.0 mL of 0.410 M HCl

H+
___ M

Cl -
___ M


(b) 15.0 mL of 0.335 M Na2SO4 and 10.6 mL of 0.200 M KCl

Na+
___ M

K+
___ M

SO42-
___ M

Cl -
___ M


(c) 3.50 g of NaCl in 46.2 mL of 0.207 M CaCl2 solution
Na+
___ M

Ca2+
___ M

Cl -
___ M

Explanation / Answer

13. A sample of solid Ca(OH)2 was stirred in water at a certain temperature until the solution contained as much dissolved Ca(OH)2 as it could hold. A90.8-mL sample of this solution was withdrawn and titrated with 0.0558 M HBr. It required 61.7 mL of the acid solution for neutralization.

(a) What was the molarity of the Ca(OH)2 solution?

Ca(OH)2 aq + 2HBr ---> 2H2O + CaBr2

1 mol of base : 2 mol of acid

calculate moles of acid used

N = M*¨V = 0.0558*61.7 = 3.443 mmol

But we require half of those to neutralize 1 mol so

3.443/2 = 1.721 mmol of base

Find molarity

M = n/V = 1.721 / 90.8 = 0.01895 M

M of Ca(OH)2 = 0.01895 M

(b) What is the solubility of Ca(OH)2 in water, at the experimental temperature, in grams of Ca(OH)2 per 100 mL of solution?

MW of Ca(OH)2 = 74.093

m = 0.01895*74.093 =1.404 g per liter

we want per 100 ml so

0.1404 g per 100 ml (0.1L)

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