Common ion effect;At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100?
Common ion effect;At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl2. Once equ?
At some other temperature, 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 20.00 mL sample requires 38.80 mL of 0.0100 M HCl to neutralize it.
Calculate the value for Ksp of Ca(OH)2 from this data.
What is the solubility of Ca(OH)2 in this solution, in g/100mL
thanks in advance:)
I worked this today but I don't remember these numbers. I wonder if I misinterpreted the first post or if this one is actually different?
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
moles HCl = M x L = 0.03880 x 0.01 =?
moles Ca(OH)2 = 1/2 that
mols Ca(OH)2 x molar mass = grams/20 mL.
grams x (100/20) = ? g/100 mL.
Ksp = (Ca^2+)(OH^-)^2
Ca^2+ = 0.01M
You have mols Ca(OH)2 from above. Calculate OH^- from that and divide by 0.02 to convert to M then substitute into Ksp expression.
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To calculate the value for Ksp of Ca(OH)2, we need to use the information provided about the neutralization of the solution and the solubility of Ca(OH)2.
1. From the given information, we know that 38.80 mL of 0.0100 M HCl is required to neutralize 20.00 mL of the Ca(OH)2 solution. This means that the moles of HCl used is equal to the moles of OH- ions present in the 20.00 mL solution.
2. Using the equation Ca(OH)2 + 2HCl -> CaCl2 + 2H2O, we can determine that each Ca(OH)2 molecule reacts with 2 moles of HCl. Therefore, the moles of OH- ions in the solution is equal to half of the moles of HCl used.
3. Calculate the moles of HCl used:
Moles of HCl = concentration of HCl x volume of HCl used
Moles of HCl = 0.0100 M x 0.0388 L = 0.000388 moles
Since the reaction is 1:2 between Ca(OH)2 and HCl, the number of moles of OH- ions is half of this value:
Moles of OH- = 0.000388 moles / 2 = 0.000194 moles
4. Next, we need to calculate the concentration of Ca2+ ions in the solution. The Ca2+ ions come from the dissociation of Ca(OH)2, and each mole of Ca(OH)2 produces one mole of Ca2+ ions.
Concentration of Ca2+ ions = Moles of Ca2+ ions / Volume of solution (in L)
The volume of the solution can be calculated by subtracting the volume of the undissolved solid Ca(OH)2 from the total volume of the solution:
Volume of solution = 20.00 mL - 0.0100 mL = 0.01990 L
Assuming the dissolution of Ca(OH)2 is complete, the concentration of Ca2+ ions is:
Concentration of Ca2+ ions = 0.000194 moles / 0.01990 L = 0.00975 M
5. Finally, we can calculate the solubility of Ca(OH)2 in this solution in g/100 mL. The solubility is simply the mass of Ca(OH)2 that dissolved in 100 mL of solution.
Using the molar mass of Ca(OH)2 (74.093 g/mol) and the concentration of Ca2+ ions, we can calculate the mass of Ca(OH)2 in 100 mL of solution:
Mass of Ca(OH)2 = Concentration of Ca2+ ions x Molar mass of Ca(OH)2 x Volume of solution (in L)
Volume of solution in 100 mL = 0.100 L
Mass of Ca(OH)2 = 0.00975 M x 74.093 g/mol x 0.100 L = 0.0726 g/100 mL
Therefore, the solubility of Ca(OH)2 in this solution is 0.0726 g/100 mL.
To find the value for Ksp of Ca(OH)2, we need to consider the equation Ca(OH)2 (s) -> Ca2+ (aq) + 2OH- (aq). Since the concentration of Ca2+ ions is 0.00975 M and the concentration of OH- ions is 0.000194 M, we can plug these values into the expression for Ksp:
Ksp = [Ca2+][OH-]^2
Ksp = (0.00975 M)(0.000194 M)^2
Calculating this value will give you the Ksp of Ca(OH)2.