I titrated a solution of 25.0 mL of a 0.050 M NaOH solution containing Ca(OH)2 u
ID: 945629 • Letter: I
Question
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.
Explanation / Answer
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)