Suppose a student titrates a 10.00-mL aliquot of saturated Ca(OH)2 solution to the equivalence point with 13.65 mL of 0.0234 M HCl.
What was the initial [OH − ]?
What is the experimental value of Ksp?
To find the initial [OH^-] and Ksp, we need to calculate the moles of Ca(OH)2 and HCl used in the titration.
Step 1: Calculate the moles of HCl used.
Given:
Volume of HCl used = 13.65 mL = 0.01365 L (converted from mL to L)
Molarity of HCl = 0.0234 M
Moles of HCl = Molarity × Volume
Moles of HCl = 0.0234 M × 0.01365 L = 0.0003196 moles
Step 2: Calculate the moles of Ca(OH)2.
Since Ca(OH)2 is a strong base, it reacts in a 1:2 ratio with HCl.
Therefore, moles of Ca(OH)2 = 2 × Moles of HCl = 2 × 0.0003196 moles = 0.0006392 moles
Step 3: Calculate the initial moles of Ca(OH)2.
Given:
Volume of Ca(OH)2 solution used = 10.00 mL = 0.01000 L (converted from mL to L)
Molarity of Ca(OH)2 can be calculated using the equation M1V1 = M2V2, where:
M1 = Initial molarity of Ca(OH)2
V1 = Initial volume of Ca(OH)2 solution
M2 = Final molarity of Ca(OH)2 at equivalence point
V2 = Volume of HCl used at the equivalence point
Moles of Ca(OH)2 = Molarity × Volume
0.0006392 moles (moles of Ca(OH)2) = M1 × 0.01000 L
Rearranging the equation and solving for M1:
M1 = 0.0006392 moles / 0.01000 L = 0.06392 M
Step 4: Calculate the initial concentration of OH^-.
Since Ca(OH)2 is a strong base and dissociates completely, the concentration of OH^- is twice the molarity of Ca(OH)2.
Concentration of OH^- = 2 × M1 (from step 3)
Concentration of OH^- = 2 × 0.06392 M = 0.1278 M
Therefore, the initial [OH^-] is 0.1278 M.
Step 5: Calculate the Ksp value.
At the equivalence point of the titration, the moles of Ca(OH)2 and HCl react in a 1:1 ratio.
Thus, moles of HCl = Moles of Ca(OH)2
Molarity of Ca(OH)2 = 0.0003196 moles (moles of HCl) / 0.01000 L (volume of Ca(OH)2 solution used)
Ksp = [Ca^2+][OH^-]^2
Since [OH^-] is twice the molarity of Ca(OH)2:
Ksp = (Molarity of Ca(OH)2) × ([OH^-])^2
Ksp = 0.0003196 moles / 0.01000 L × (0.1278 M)^2
Calculating: Ksp ≈ 0.000649
Therefore, the experimental value of Ksp is approximately 0.000649.
To find the initial [OH^-] in the saturated Ca(OH)2 solution, we can use the balanced chemical equation for the reaction between Ca(OH)2 and HCl:
Ca(OH)2 + 2HCl -> CaCl2 + 2H2O
From this equation, we can see that 1 mole of Ca(OH)2 reacts with 2 moles of HCl. Therefore, the number of moles of HCl used in the titration is:
moles of HCl = volume of HCl (in L) * molarity of HCl
moles of HCl = 0.01365 L * 0.0234 mol/L = 0.00031941 mol
Since 1 mole of Ca(OH)2 reacts with 2 moles of HCl, the number of moles of Ca(OH)2 in the 10.00 mL aliquot can be calculated:
moles of Ca(OH)2 = 0.00031941 mol/2 = 0.000159705 mol
The volume of the 10.00 mL aliquot in liters is 0.01000 L. Therefore, the initial concentration of OH^- can be calculated as follows:
[OH^-] = moles of Ca(OH)2 / volume of the aliquot (in L)
[OH^-] = 0.000159705 mol / 0.01000 L
Now, to find the experimental value of Ksp, we need to determine the amount of Ca(OH)2 that dissolved in the solution. At the equivalence point, all of the Ca(OH)2 had reacted with HCl. Therefore, the moles of Ca(OH)2 dissolved is equal to the moles of HCl used in the titration:
moles of Ca(OH)2 dissolved = 0.00031941 mol
The molar mass of Ca(OH)2 is 74.093 g/mol, so we can calculate the mass of Ca(OH)2 dissolved:
mass of Ca(OH)2 = moles of Ca(OH)2 dissolved * molar mass of Ca(OH)2
mass of Ca(OH)2 = 0.00031941 mol * 74.093 g/mol
Ksp is the solubility product constant for Ca(OH)2. It can be calculated using the formula:
Ksp = [Ca2+] * [OH^-]^2
Since Ca(OH)2 dissolves completely in water, the concentration of Ca2+ is twice the concentration of OH^-.
Therefore, Ksp = 4 * [OH^-]^3
To find the experimental value of Ksp, we substitute the concentration of OH^- from the initial calculation and calculate:
Ksp = 4 * ([OH^-])^3
Now you have the initial [OH^-] and the experimental value of Ksp for the saturated Ca(OH)2 solution titrated with HCl.
First calc. the moles from M and L: (.0234)(.01365)= 3.19e-4
Determine moles Ca(OH)2 titrated:
(3.19e-4)/2 = 1.60e-4 mol
Remember, every one Ca(OH)2 titrated requires 2 H+
[Ca2+] = (1.60e-4)/.01 = .01597M
[OH-] = (.01597)(2) = .0319 M
Ksp = (.01597)(.0319)^2 = 1.63e-5