27.39 g please finish the table sample1 sample 2 Sample 3 mass of flask 24.3 g 2
ID: 1045878 • Letter: 2
Question
27.39 g
please finish the table
sample1 sample 2 Sample 3 mass of flask 24.3 g 24.32 g 24.33 g mass of flask+lime water 27.29 g27.39 g
27.38 g mass of lime water 2.99 g 3.07 g 3.05 g volume Ca(OH)2 .334 ml .325 ml .327 ml Concentration HCl (M) 0.1 M 0.1 M 0.1 M initial HCl volume 1 ml 1 ml 1 ml Final HCl volume .95 ml .96 ml .95 ml HCl delivered .05 ml .04 ml .05 ml Moles HCl delivered 2.04X10-3 1.63X10-3 2.04X10-3 Moles OH in sample Moles Ca+ in sample Molar solubility calculated Ksp Average KspExplanation / Answer
Sample 1
Sample 2
Sample 3
Mass of flask (g)
24.30
24.32
24.33
Mass of flask and lime water (g)
27.29
27.29
27.38
Mass of lime water (g)
2.99
2.97
3.05
Volume of Ca(OH)2 (mL)
0.334
0.325
0.327
Concentration HCl (M)
0.1
0.1
0.1
Initial HCl volume (mL)
1
1
1
Final HCl volume (mL)
0.95
0.95
0.95
HCl delivered (mL)
0.05
0.05
0.05
Moles HCl delivered (mol) = (volume of HCl in L)*(concentration of HCl)
(0.05 mL)*(1 L/1000 mL)*(0.1 M) = 5.0*10-6
(0.05 mL)*(1 L/1000 mL)*(0.1 M) = 5.0*10-6
(0.05 mL)*(1 L/1000 mL)*(0.1 M) = 5.0*10-6
Moles OH- in sample (mol) (calculation 1)
5.0*10-6
5.0*10-6
5.0*10-6
Moles Ca2+ in sample (mol) (calculation 2)
2.5*10-6
2.5*10-6
2.5*10-6
Molar solubility (M) (calculation 3)
7.48*10-3
7.69*10-3
7.64*10-3
Calculated Ksp (calculation 4)
1.67*10-6
1.82*10-6
1.78*10-6
Average Ksp
1/3*(1.67*10-6 + 1.82*10-6 + 1.78*10-6) = 1.76*10-6
Calculation 1:
HCl reacts with Ca(OH)2 as below.
2 HCl (aq) + Ca(OH)2 (aq) ------> CaCl2 (aq) + 2 H2O (l)
The net ionic equation for the reaction is
H+ (aq) + OH- (aq) ------> H2O (l)
As per the stoichiometric equation,
1 mole H+ = 1 mole OH-
Therefore,
5.0*10-6 mole HCl = 5.0*10-6 mole OH-
Calculation 2:
Consider the ionization of Ca(OH)2 as below.
Ca(OH)2 (aq) ------> Ca2+ (aq) + 2 OH- (aq)
As per the stoichiometric equation,
2 moles OH- = 1 mole Ca2+.
Therefore,
5.0*10-6 mole OH- = ½*(5.0*10-6) mole = 2.5*10-6 mole Ca2+
Calculation 3:
Molar solubility of Ca(OH)2 = molar solubility of Ca2+ = (moles of Ca2+)/(volume of solution in L).
Take sample 1 as an example.
Molar solubility of Ca(OH)2 = (2.5*10-6 mole)/[(0.334 mL)*(1 L/1000 mL)] = 7.48*10-3 M.
Calculation 4:
Consider the ionization of solid Ca(OH)2.
Ca(OH)2 (s) <======> Ca2+ (aq) + 2 OH- (aq)
Molar solubility of OH- = 2*(molar solubility of Ca2+)
Take sample 1 as an example.
Molar solubility of OH- = 2*(7.48*10-3 M) = 0.01496 M.
Ksp = [Ca2+][OH-]2 = (7.48*10-3)*(0.01496)2 = 1.67*10-6 (no unit)
Sample 1
Sample 2
Sample 3
Mass of flask (g)
24.30
24.32
24.33
Mass of flask and lime water (g)
27.29
27.29
27.38
Mass of lime water (g)
2.99
2.97
3.05
Volume of Ca(OH)2 (mL)
0.334
0.325
0.327
Concentration HCl (M)
0.1
0.1
0.1
Initial HCl volume (mL)
1
1
1
Final HCl volume (mL)
0.95
0.95
0.95
HCl delivered (mL)
0.05
0.05
0.05
Moles HCl delivered (mol) = (volume of HCl in L)*(concentration of HCl)
(0.05 mL)*(1 L/1000 mL)*(0.1 M) = 5.0*10-6
(0.05 mL)*(1 L/1000 mL)*(0.1 M) = 5.0*10-6
(0.05 mL)*(1 L/1000 mL)*(0.1 M) = 5.0*10-6
Moles OH- in sample (mol) (calculation 1)
5.0*10-6
5.0*10-6
5.0*10-6
Moles Ca2+ in sample (mol) (calculation 2)
2.5*10-6
2.5*10-6
2.5*10-6
Molar solubility (M) (calculation 3)
7.48*10-3
7.69*10-3
7.64*10-3
Calculated Ksp (calculation 4)
1.67*10-6
1.82*10-6
1.78*10-6
Average Ksp
1/3*(1.67*10-6 + 1.82*10-6 + 1.78*10-6) = 1.76*10-6
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