Each trail we started with 15 mL of Ca(OH)2 We used 0.050 M HCl for the tritatio

Question

Each trail we started with 15 mL of Ca(OH)2

We used 0.050 M HCl for the tritation

Trial 1 Equivalence point: 9.6 mL

Trial 2 Equivalence point: 12.6 mL

Calculate the [OH–] from the results of your titrations. Explain your calculations.

2.   Calculate the [Ca2+]. Explain your calculations.

3.   Calculate the Ksp for calcium hydroxide. Explain your calculations.

4.   Find the accepted value of the Ksp for calcium hydroxide and compare it with your value. Discuss the discrepancy and suggest possible sources of experimental error.

Explanation / Answer

Each trail we started with 15.0 mL of Ca(OH)2

We used 0.050 M HCl for the Titration &

Trial 1 Equivalence point: 9.6 mL and Trial 2 Equivalence point: 12.6 mL

Average Volume for Equivalence point = (9.6+ 12.6)/2 = 11.1 ml

The overall reaction for this titration would be:

Ca(OH)2 (aq) + 2 HCl (aq)  ----------> CaCl2 (aq)+ 2 H2O (aq)

1. Calculate the [OH–] from the results of your titrations. Explain your calculations.

To calculate [OH–], we need to calculate moles of  [OH–], as  [OH–] = Moles of [OH–]/ volume of [OH–] in L.

Foe moles of [OH–], we need to calculate moles of HCl.

moles of HCl = molarity of HCl * volume of HCl in L

moles of HCl = 0.050 moles/L * 11.1 x 10-3 L = 5.55 x 10-4 moles of HCl

Moles of OH- = 5.55 x 10-4 moles of HCl * (1 mole Ca(OH)2 /2 moles HCl ) * (2 moles OH- / 1 mole Ca(OH)2 )

So Moles of OH- = 5.55 x 10-4 moles

Total volume used = 15.0 ml = 15.0 x 10-3 L

[OH–] = moles of OH- / vol in L = 5.55 x 10-4 moles / 15.0 x 10-3 L

[OH–] = 0.0370 M

2.   Calculate the [Ca2+]. Explain your calculations.

[Ca2+] = moles of [Ca2+]/vol in L

Moles of Ca2+]= moles of OH- * (1 mole Ca(OH)2/ 2 moles of OH- ) * (1 mole of Ca2+ / 1 mole Ca(OH)2 )

Moles of Ca2+ = 5.55 x 10-4 moles of OH- * (1 mole Ca(OH)2/ 2 moles of OH- ) * (1 mole of Ca2+ / 1 mole Ca(OH)2 )

Moles of Ca2+ = 2.78 x 10-4 moles of Ca2+

Total Volume = 15.0 x 10-3 L

[Ca2+] = 2.78 x 10-4 moles / 15.0 x 10-3 L = 0.0185 M

3.   Calculate the Ksp for calcium hydroxide. Explain your calculations.

Ca(OH)2 (s) ----------> Ca2+(aq)+ 2OH- (aq)

Ksp = {[Ca2+] [OH-]2}

The concentrations of [Ca2+] and [OH-] calculated above are at equilibrium, so

Ksp =  {[Ca2+] [OH-]2} = 0.0185 x (0.0370)2 = 2.53 x 10-5

4.   Find the accepted value of the Ksp for calcium hydroxide and compare it with your value. Discuss the discrepancy and suggest possible sources of experimental error.

The accepted value of Ksp for calcium hydroxide is 6.0 x 10-6 and the value obtained through the above experiment is 25.3 x 10-6

So the % error = Mod(actual - observed)/actual * 100 % = Mod(25.3 x 10-6 - 6.0 x 10-6 )/ 6.0 x 10 -6 * 100 % = 322 %

The error can be due to many reasons. It can be due to any acid or base leftover on the sides of the flask, or at the tip of the burette or an air bubble in the burette, measurement errors, etc.

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