An experiment was performed to determine the K_sp of Ca(OH)_2. A solution contai

Question

An experiment was performed to determine the K_sp of Ca(OH)_2. A solution containing 2.00 times 10^-2 M Ca(NO_3)_2 was saturated with Ca(OH)_2 and filtered. Two samples of the saturated solution were titrated with 5.021 times 10^-2 M HCI, in order to determine the OH^- ion concentration in the saturated solution. In doing this problem, note that 1 mol of HCI reacts with 1 mol of OH^- ions. This relationship is the basis for the determination of the OH^- ion concentration: 1 mol Ca^2 + ions are produced for every 2 mols of OH^- ions produced. The data resulting from this experiment are shown in Table 2.

Explanation / Answer

Determination 1 :

(a)Moles of OH- titrated = moles of HCl added = 5.021*10^-2M* 16.44mL/1000mL = 0.083*10^-2 Moles

(b) [OH-]saturated solution = 0.083*10^-2 Moles *1000mL/(52.2+16.44)mL =0.012 M

(c) inital concentration of Ca^2+ is the concentration due to Ca(NO3)2

[Ca^2+] initial = 2.00*10^-2 M

(d) Ca(OH)2 <==> Ca^2+ + 2OH-

2 moles of OH- is produced for 1 mole of Ca^2+.

[Ca^2+] from ca(OH)2 = 0.006 M

(e) [Ca^2+] in saturated solution = [Ca^2+] from Ca(NO3)2 + [Ca^2+] from ca(OH)2

moles of Ca^2+ from Ca(NO3)2 =2*10^-2 M *0.0522 L =1.044*10^-3 moles

moles of Ca^2+ from Ca(OH)2 = 0.083*10^-2 Moles/2 =0.0415*10^-2 moles

total Ca^2 IN SATURATED SOLUTION = 1.044*10^-3 + 0.0415*10^-2 =0.0015 moles

molarity = 0.0015 moles/total volume = 0.0015/52.2 = 0.028 M

(f) Ksp = [Ca^2+][OH-]^2

=0.028* (0.012)^2

=4.02*10^-6

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Determination 2

(a)Moles of OH- titrated = moles of HCl added = 5.021*10^-2M* 15.49mL/1000mL = 0.078*10^-2 Moles

(b) [OH-]saturated solution = 0.078*10^-2 Moles *1000mL/(48.7+15.49)mL =0.012 M

(c) inital concentration of Ca^2+ is the concentration due to Ca(NO3)2

[Ca^2+] initial = 2.00*10^-2 M

(d) Ca(OH)2 <==> Ca^2+ + 2OH-

2 moles of OH- is produced for 1 mole of Ca^2+.

[Ca^2+] from ca(OH)2 = 0.006 M

(e) [Ca^2+] in saturated solution = [Ca^2+] from Ca(NO3)2 + [Ca^2+] from ca(OH)2

moles of Ca^2+ from Ca(NO3)2 =2*10^-2 M *0.0522 L =1.044*10^-3 moles

moles of Ca^2+ from Ca(OH)2 = 0.078*10^-2 Moles/2 =0.039*10^-2 moles

total Ca^2 IN SATURATED SOLUTION = 1.044*10^-3 + 0.039*10^-2 =0.00143 moles

molarity = 0.00143 moles/total volume = 0.0015/0.0487 L = 0.029 M

(f) Ksp = [Ca^2+][OH-]^2

=0.029* (0.012)^2

=4.17*10^-6

Average = 4.095*10^-6

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