Question

Chem experiment:

Data:

initial burette reading(ml): final burette reading: volume titrant added:

trial 1: 32.00ml 48.0ml 16ml

trial 2: 32.00ml 43.8ml 11.8ml

trial 3: 34.00ml 43.9ml 9.9ml

the volume of HCl (titrant) was == 1.26 mili-moles

Complete the following ICE tables and use them to calculate the value of Ksp for all three trials. The initial concentrations of Ca2+ and OH– refer to the concentrations before any of the solid dissolved (i.e.when the solutions were originally made), and can therefore be assumed to be zero

Table 1: ICE table for Trial 1

Average	ICE	[Ca(OH)2] (M)	[Ca2+] (M)	[OH–] (M)
Concentration (M)	I
	C
	E

K_sp =

Table 2: ICE table for Trial 2

Average	ICE	[Ca(OH)2] (M)	[Ca2+] (M)	[OH–] (M)
Concentration (M)	I
	C
	E

K_sp =

Table 3: ICE table for Trial 3

Average	ICE	[Ca(OH)2] (M)	[Ca2+] (M)	[OH–] (M)
Concentration (M)	I
	C
	E

K_sp =

1. Calculate the average of the equilibrium/solubility product constant using all runs.

K_sp =

The literature value for K_sp from your textbook is reported as 4.68 x 10^-6 (Tro, Principles of Chemistry). Calculate the percent error in your K_sp value using the formula below. Explain any discrepancies between your results

Assume in a different experiment, you attempt to dissolve 0.55 grams of calcium hydroxide into 500.0 mL of water. Will all of the solid dissolve? (Hint: Calculate the concentrations of Ca²⁺ and OH^– this would produce and use this to calculate Q; then compare that to K_sp.)

What were the main results found in this experiment? Include both quantitative results (i.e. important calculated values, including percent error) and qualitative results (i.e. important properties that were investigated)

UPDATE!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Fill the burette with ~10 mL of the standardized 0.100 M HCl solution; drain and refill it with ~15 mL

of 0.100 M HCl solution

.00126 moles H⁺

Answer 1

I think you have to dissolve calcium hydroxide in the water and titrated the filtrate with acid HCl. HCl has molarity of 0.1M

Let's say x gram of Ca(OH)2 dissolve in 500 ml of water. So the molarMol of Ca(OH)2 = x/ 74.093 × 1000/500 = x/37.0465 M, [Ca2+]= x/37.0465 M, [OH-] = x/18.52325 M, moles of Ca2+ = x/ 74.093 in 500 ml of water, moles of OH- = x/37.0465 in 500 ml of water.

Table 1: ICE table for Trial 1, volume of HCl = 16.0 ml = 0.0016 moles of HCl

Average	ICE	[Ca(OH)2] (M)	[Ca2+] (M)	[OH–] (M)
Concentration (M)	I		x/37.0465	x/18.52325
	C		0	- 0.0032
			x/ 37.0465	0

x = 0.0032 × 18.52325 = 0.05927 g of Ca(OH)2 dissolve

Ksp = 0.0016 × (0.0032)² = 1.6384 × 10^-8

Similarly you can do for other trial.