Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 60.00 mL sample titrated with HCl requires 11.13 mL of 9.83×10−2 M HCl to reach the endpoint. calculate ksp for Ca(OH)2?

Question

Wyatt Atkins · Accepted Answer

undefined To calculate the $K_{sp}$ for $Ca(OH)_2$, we follow these steps:

1. **Determine the amount of $HCl$ used**: $11.13 \, \text{mL} \times 9.83 \times 10^{-2} \, \text{mol/L} = 1.094 \times 10^{-3} \, \text{mol} \, HCl$

2. **Write the neutralization reaction**: $Ca(OH)_2 + 2HCl \rightarrow CaCl_2 + 2H_2O$. This shows that 2 moles of $HCl$ react with 1 mole of $Ca(OH)_2$.

3. **Calculate moles of $Ca(OH)_2$ in the sampled solution**: Since it takes 2 moles of $HCl$ to neutralize 1 mole of $Ca(OH)_2$, the moles of $Ca(OH)_2$ is half the moles of $HCl$, which is $\frac{1.094 \times 10^{-3}}{2} = 5.47 \times 10^{-4} \, \text{mol}$.

4. **Calculate the concentration of $Ca(OH)_2$ in the solution**: The volume of the sample is 60.00 mL or 0.06000 L. So, the concentration of $Ca(OH)_2$ is $\frac{5.47 \times 10^{-4} \, \text{mol}}{0.06000 \, \text{L}} = 9.12 \times 10^{-3} \, \text{M}$.

5. **Write the dissociation equation and expression for $K_{sp}$**: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^{-}(aq)$. The $K_{sp}$ expression is $K_{sp} = [Ca^{2+}][OH^-]^2$.

6. **Calculate the ion concentrations in the saturated solution**: Since the concentration of $Ca(OH)_2$ in solution is $9.12 \times 10^{-3} \, \text{M}$, and each formula unit of $Ca(OH)_2$ produces one $Ca^{2+}$ ion and two $OH^{-}$ ions, the concentrations are:
   - $[Ca^{2+}] = 9.12 \times 10^{-3} \, \text{M}$
   - $[OH^{-}] = 2 \times 9.12 \times 10^{-3} \, \text{M} = 1.824 \times 10^{-2} \, \text{M}$

7. **Calculate $K_{sp}$**: Using the concentrations found,
   $$
   K_{sp} = (9.12 \times 10^{-3})(1.824 \times 10^{-2})^2 = 9.12 \times 10^{-3} \times 3.327 \times 10^{-4} = 3.03 \times 10^{-6}
   $$

Therefore, the $K_{sp}$ for $Ca(OH)_2$ is $3.03 \times 10^{-6}$.

Peyton Adams · Answer

#K_"sp"=3.03xx10^-6#

We work out the moles of hydroxide....

#Ca(OH)_2(aq) + 2HCl(aq) rarr CaCl_2(aq) + 2H_2O(l)#

#"Moles of HCl"=11.13*mLxx10^-3*L*mL^-1xx9.83xx10^-2*mol*L^-1=0.001094*mol#...and given this molar quantity there were HALF this molar quantity with respect to #Ca(OH)_2# in the starting #60.00*mL# aliquot....

And so #[Ca(OH)_2(aq)]=(1/2xx0.001094*mol)/(60.00*mLxx10^-3*L*mL^-1)# #=9.12xx10^-3*mol*L^-1#...

But this concentration derived from SOLUBLE #Ca(OH)_2#, the which had participated in the solubility equilibrium...

#Ca(OH)_2(s) rightleftharpoonsCa^(2+) + 2HO^-#...

for which we write the solubility expression....

#K_"sp"=[Ca^(2+)][HO^-]^2#..and we plug in the concentrations...

#=(9.12xx10^-3)xx(2xx9.12xx10^-3)^2#

#=4xx(9.12xx10^-3)^3=3.03xx10^-6#

Please check my arithmetic....no money-back guarantees...