### Calculate the Ksp of a Saturated Solution When Given Titration Data

Example #1: A solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl2. Once equilibrated, some solid Ca(OH)2 remains undissolved. The solution is filtered and a 25.00 mL sample requires 22.50 mL of 0.0250 M HCl to neutralize it. Calculate the value for Ksp of Ca(OH)2 from this data.

Solution:

1) The chemical equation:

Ca(OH)2 ⇌ Ca2+ + 2OH¯

2) The Ksp expression:

Ksp = [Ca2+] [OH¯]2

3) Use titration data to determine moles of OH¯ in the 25.0 mL sample (Remember, every one H+ neutralizes one OH¯.):

molarity = moles ÷ volume (in liters)

0.0250 mol/L = x ÷ 0.02250 L

x = 0.0005625 mol

4) Use moles of OH¯ and sample volume to determine [OH¯]:

0.0005625 mol / 0.02500 L = 0.0225 mol/L

5) Determine [Ca2+]:

it is exactly half the [OH¯]. This is because of the 1:2 molar ratio from the balanced equation.

6) Calculate the Ksp for Ca(OH)2

Ksp = (0.01125) (0.0225)2 = 5.70 x 10¯6

Example #2: 25.00 mL of saturated calcium hydroxide solution was titrated. It was found that it reacted completely with 8.13 mL of 0.102 mol L HCl. (a) Determine the solubility of Ca(OH)2 in grams per liter. (b) Determine the Ksp of Ca(OH)2

Solution to (a):

1) Determine moles of HCl used:

moles HCl = (0.102 mol / L) (0.00813 L) = 0.00082926 mol

2) Determine moles Ca(OH)2 titrated:

0.00082926 mol / 2 = 0.00041463 mol

Remember, every one Ca(OH)2 titrated requires 2 H+

3) Convert moles Ca(OH)2 to grams Ca(OH)2:

0.00041463 mol times 74.0918 g/mol = 0.030721 g

This is grams per 25.0 mL

3) Convert to grams per liter:

0.030721 g / 0.0250 L = 1.23 g/L (to three sig fig)

Solution to (b):

1) 0.00041463 mol of Ca(OH)2 in 25.0 mL means:

[Ca2+] = 0.00041463 mol / 0.0250 L = 0.0165852 M
[OH¯] = 0.0165852 M times 2 = 0.0331704 M

2) Calculate the Ksp:

Ksp = (0.0165852) (0.0331704)2 = 1.82 x 10¯5

This value is in error. The book value is 5.02 x 10¯6. The main reason for the error is that the above is actual data gathered by a high school student doing this experiment for the first time. However, the calculational technique is correct.

Example #3: A saturated solution of Pb(OH)2 is filtered and 25.00 mL of this solution is titrated with 0.000050 M HCl. The volume required to reach the equivalence point of this solution is 6.70 mL. Calculate the concentration of OH¯, Pb2+ and the Ksp of this satured solution.

Solution:

1) Determine moles of HCl used:

moles HCl = (0.000050 mol / L) (0.00670 L) = 0.000000335 mol

2) Determine moles Pb(OH)2 titrated:

0.000000335 mol / 2 = 0.0000001675 mol

Remember, every one Pb(OH)2 titrated requires 2 H+

3) 0.0000001675 mol of Pb(OH)2 in 25.0 mL means:

[Pb2+] = 0.0000001675 mol / 0.0250 L = 0.0000067 M
[OH¯] = 0.0000067 M times 2 = 0.0000134 M

4) Calculate the Ksp:

Ksp = (0.0000067) (0.0000134)2 = 1.20 x 10¯15