The solubility of slaked lime, Ca(OH)2, in water is 0.185 g/100.0 mL. Calculate the number of moles of Ca(OH)2 in 1.10×101 mL of a saturated solution.?

Question

The solubility of slaked lime, Ca(OH)2, in water is 0.185 g/100.0 mL. Calculate the number of moles of Ca(OH)2 in 1.10×101 mL of a saturated solution.?

1 Answer

Impact of this question

9213 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-05T12:08:05

In what volume....?

Explanation:

We interrogate the equilibrium....

$C a {(O H)}_{2} (s) H_{2} O - - \to C a^{2 +} + 2 H O^{-}$ $Ca(OH)_2(s) stackrel(H_2O)rarrCa^(2+) + 2HO^-$

For which, $K_{sp} = [C a^{2 +}] {[H O^{-}]}^{2} = 5.5 \times 10^{- 6}$ $K_"sp"=[Ca^(2+)][HO^-]^2=5.5xx10^-6$ according to [this site...](https://en.wikipedia.org/wiki/Calcium_hydroxide)

If we put $solubility of calcium hydroxide \equiv S$ $"solubility of calcium hydroxide"-=S$ , then substituting in the $K_{sp}$ $K_"sp"$ expression....we get....

$K_{sp} = S \times {(2 S)}^{2} = 4 S^{3}$ $K_"sp"=Sxx(2S)^2=4S^3$ ....and thus $S =^{3} \sqrt{\frac{K_{sp}}{4}}$ $S=""^(3)sqrt((K_"sp")/(4))$

$=^{3} \sqrt{\frac{5.5 \times 10^{- 6}}{4}} = 0.0111 \cdot m o l \cdot L^{- 1}$ $=""^3sqrt((5.5xx10^-6)/(4))=0.0111*mol*L^-1$

A gram solubility of ...........................

$0.0111 \cdot m o l \cdot L^{- 1} \times 74.09 \cdot g \cdot m o l^{- 1} = 0.824 \cdot g \cdot L^{- 1}$ $0.0111*mol*L^-1xx74.09*g*mol^-1=0.824*g*L^-1$ ...which is different to the solubility you quoted. I wonder who has the correct value.

Science

Math

Humanities

... and beyond

The solubility of slaked lime, Ca(OH)2, in water is 0.185 g/100.0 mL. Calculate the number of moles of Ca(OH)2 in 1.10×101 mL of a saturated solution.?

1 Answer

Explanation:

Related questions

Impact of this question