2. (7pts total) Answer the following questions based on the sparing soluble salt

Question

2. (7pts total) Answer the following questions based on the sparing soluble salt, Ca(OH)2. Be sure to include proper significant figures and units when needed The class average Kp for Ca(OH)2 (as measured by titration) was 2.9x10+/-0.5x10. This is in reasonable agreement with the value reported in your text book (5.5x10). However, there are other sources of information regarding the solubility of Ca(OH)2 that may be more reliable than your te book. The National Lime Association (yes, that is a real thing/) has published a "Fact Sheet" listing many physical properties of Lime, or Ca(OH)2, including: 1) the solubility of Ca(OH)2 (reported in units of grams per 100.000 grams of sa as a function of temperature and 2) the pH of saturated Ca(OH)2 solutions as a function of temperature. ion can be used to calculate the Kip of Ca(0H) for comparison with the class average results and the text book value of 5.5x1o Follow the prompts to calculate the K, of Ca(OH)2 at 25C using the National Lime Association reported solubility of 0.159 grams of Ca(OH) per 100.000 grams of 25 °C saturated solution. a) (1pt) How many moles of CalOHl2 are present 100.000 grams of 25 °C saturated soln? The solubility was reported as 0.159 grams of Ca(OH) in 100.00 grams of 25 °C saturated soln. The molar mass of Ca(OH)2 is 74.09 g/mol. (1pt) How many gramsofHL are present in 100.000 grams of 25°C saturated soln? Hint: the soln contains only water and Ca(OH),2 (1pt) Use the grams of H2O calculated above and a density of 0.997044 g/mL reported for 25°C water to calculate the volume ofHao (in L) in 25°C saturated son. (1pt) The volume occupied by 0.159 grams of Ca(OH) is assumed to be negligible compared to the volume of water, so you can use the moles of Ca(0H)h and the volume of water, both calculated above, to calculate the molarity ofCalOHk in a saturated soln. at 25°C. (1pt) Use the molarity calculated above as "s" in the Kp expression to calculate to calculate the Kg of Ca(OH) at 25 °C.

Explanation / Answer

Ksp for Ca(OH)2

a) 0.159 g Ca(OH)2 in 100 g solution

moles of Ca(OH)2 = 0.159 g/74.09 g/mol = 0.00215 mol

grams of water = 100 - 0.159 = 99.841 g

volume of water = 99.841 g/0.997044 g/ml = 100.137 ml

molarity of Ca(OH)2 solution = 0.00215 mol/0.100317 L = 0.0214 M

Ksp of Ca(OH)2 = [Ca2+][OH-]^2 = (0.0214)^3 = 9.80 x 10^-6

b) No, the two Ksp values does not match

The Ksp calculated is less than the Ksp reported

Ca(OH)2 is less soluble at high temperature than it is at low temperature.

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