A sample of ca(oh)2 is allowed to stand in distilled water until a saturated solution forms. A 50.00 mL sample of this saturated solution is neutralized with 24.4 mL of 0.05 mol/L Hcl
2hcl+ ca(oh)2>cacl2+2h2o
Use this information to calculate the ksp for ca(oh)2
mols HCl used = M x L = ?
mols Ca(OH)2 = 1/2 that from the coefficients in the balanced equation (of the titrated sample which is 50 mL).
mols Ca(OH)2 in the original sample of 0.05L. Convert to mols/L, substitute into Ksp expression and solve for Ksp.
I don't get it could you explain it with the calculations
So I tell you HOW to do it and you ask me to do your work for you?
mols HCl used = M x L = 0.0244L*0.050M = 0.00122 mols
mols Ca(OH)2 = 1/2 that from the coefficients in the balanced equation (of the titrated sample which is 50 mL).
mols Ca(OH)2 in the original sample of 0.05L.0.00122*1/2 = 0.00061 mols Ca(OH)2 in the 50 mL.
Convert to mols/L, substitute into Ksp expression and solve for Ksp.
0.00061*(1000 mL/50 mL) = 0.0122M = [Ca(OH)2]. If [Ca(OH)2] = 0.0122 then (Ca^2+) = 0.0122 and (OH^-) = 2*0.0122 = 0.0244 and
Ksp = (Ca^2+)(OH^-)^2 =
(0.0122)(0.0244)^2 = ?
Check my work carefully.
To calculate the Ksp (solubility product constant) for Ca(OH)2, we need to use the balanced chemical equation and the information provided.
The balanced chemical equation is:
2HCl + Ca(OH)2 → CaCl2 + 2H2O
From the equation, we can see that the molar ratio between HCl and Ca(OH)2 is 2:1.
Given that 24.4 mL of 0.05 mol/L HCl was used, we can calculate the moles of HCl used:
moles of HCl = concentration of HCl (mol/L) × volume of HCl (L)
moles of HCl = 0.05 mol/L × 0.0244 L
moles of HCl = 0.00122 mol
Since the molar ratio between HCl and Ca(OH)2 is 2:1, the moles of Ca(OH)2 in the saturated solution are twice the moles of HCl used:
moles of Ca(OH)2 = 2 × 0.00122 mol
moles of Ca(OH)2 = 0.00244 mol
Now, we need to calculate the concentration of Ca(OH)2 in the saturated solution. We know that 50.00 mL of the saturated solution was used, so the volume in liters is:
volume of saturated solution (L) = 50.00 mL ÷ 1000 mL/L
volume of saturated solution (L) = 0.0500 L
The concentration of Ca(OH)2 is:
concentration of Ca(OH)2 (mol/L) = moles of Ca(OH)2 ÷ volume of saturated solution (L)
concentration of Ca(OH)2 (mol/L) = 0.00244 mol ÷ 0.0500 L
concentration of Ca(OH)2 (mol/L) = 0.0488 mol/L
The expression for the solubility product constant (Ksp) is obtained from the balanced chemical equation with the stoichiometric coefficients as exponents:
Ksp = [Ca2+][OH-]^2
Since the molar ratio between Ca(OH)2 and Ca2+ is 1:1 and the concentration of Ca(OH)2 is the same as the concentration of Ca2+, we can substitute the concentration of Ca(OH)2 into the expression:
Ksp = (concentration of Ca(OH)2)^2 × (concentration of OH-)^2
Since Ca(OH)2 is a strong base, it completely dissociates to produce two OH- ions: Ca(OH)2 → Ca2+ + 2OH-
Therefore, the concentration of OH- is twice the concentration of Ca(OH)2:
concentration of OH- = 2 × concentration of Ca(OH)2
concentration of OH- = 2 × 0.0488 mol/L
Now, we can substitute the values into the equation to calculate the Ksp for Ca(OH)2:
Ksp = (0.0488 mol/L)^2 × (2 × 0.0488 mol/L)^2
Ksp = 0.000235 mol^3/L^3
Therefore, the calculated Ksp for Ca(OH)2 is 0.000235 mol^3/L^3.