Question

# I titrated a solution of 25.0 mL of a 0.050 M NaOH solution containing Ca(OH)2 using...

I titrated a solution of 25.0 mL of a 0.050 M NaOH solution containing Ca(OH)2 using 27.71 mL of standardized 0.065 M HCl.

1.) use the HCl volume and concentration to calculate [OH- ] in the saturated solution of Ca(OH)2 in NaOH solution.

2.) Using the known concentration of NaOH solution in Part B, calculate [OH- ] due to the NaOH alone.

3.) Calculate [OH- ] due to dissolved Ca(OH)2.

4.) Calculate [Ca^2+ ].

5.) Calculate Ksp for Ca(OH)2 in the saturated solution of Ca(OH)2 in NaOH.

6.) Calculate the solubility (in g/L) for Ca(OH)2 in the NaOH solution.

1. The moles of OH- in the saturated solution of Ca(OH)2 in NaOH is equal to the moles of HCl used.

Moles of OH- = MxV = 0.065 M x 0.02771 L = 0.001801 mol

Hence concentration of OH-, [OH-] = 0.001801 mol / 0.025L = 0.0720 M (answer)

2. [OH-] due to NaOH alone = MxV = 0.050 M x 0.025 L / 0.025 L = 0.050 M (answer)

3. [OH-] due to dissolved Ca(OH)2 = 0.0720 - 0.050 = 0.022 M (answer)

4. Ca(OH)2 <---- > Ca2+(aq) + 2 OH-(aq)

1 mol-------------- 1 mol, ----- 2 mol

[Ca2+(aq)] =  0.022 M / 2 = 0.011 M (answer)

5. Ksp = [Ca2+(aq)] x [OH-(aq)]2  = (0.011 M) x(0.022M)2 = 5.32x10-6 (answer)