Ksp of calcium hydroxide [closed]








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What would the Ksp of calcium hydroxide be at around room temp (66 Fahrenheit) and 1 atm of pressure? I know the Ksp value varies with temperature, but is there a notable difference from 20 degrees C to 25 degrees C? 




According to [IUPAC Solubility Data Series Volume 52, *Alkaline Earth Hydroxides in Water and Aqueous Solutions* (1992)](http://dx.doi.org/10.1016/B978-0-08-023920-0.50010-1), the best fitting to experimental solubility values of calcium hydroxide in water was obtained with a three-parameter equation:


$\ln\left({b/{\text{mol kg}}^{-1}}\right) = 86.1534 - 3492.14/({T/{\text{K}}}) - 13.7494\,\ln({T/{\text{K}}})$


where $b$ is molality and $T$ is temperature.


The results have a standard uncertainty of


$\sigma \left( b \right) = 1.7 \times {10^{ - 4}}\;{\text{mol k}}{{\text{g}}^{ - 1}}$ 


over the temperature range from 273.15 K to 623.15 K.




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For


${T\_1} = 20\;^\circ\text{C} = 293.15\;{\text{K}}$


the three-parameter equation reads


$\begin{aligned}
 \ln\left({{b\_1}/{\text{mol kg}^{-1}}}\right) = & \;86.1534 - 3492.14/({T\_1}/{\text{K)}} - 13.7494\,\ln({T\_1}/{\text{K)}} \\ 
 = & \;86.1534 - 3492.14/(293.15\;{\text{K}}/{\text{K)}} - 13.7494\,\ln(293.15\;{\text{K}}/{\text{K)}} \\ 
 = & \; {-}3.8651 \\ 
 {b\_1}/{\text{mol kg}^{-1}} = & \;0.02096 \\ 
 {b\_1} = & \;0.02096\;{\text{mol kg}^{-1}} \\ 
\end{aligned} $


In the same way for


${T\_2} = 25\;^\circ\text{C} = 298.15\;{\text{K}}$


the three-parameter equation reads


$\begin{aligned}
 \ln\left({{b\_2}/{\text{mol kg}^{-1}}}\right) = & \;86.1534 - 3492.14/({T\_2}/{\text{K)}} - 13.7494\,\ln({T\_2}/{\text{K)}} \\ 
 = & \;86.1534 - 3492.14/(298.15\;{\text{K}}/{\text{K)}} - 13.7494\,\ln(298.15\;{\text{K}}/{\text{K)}} \\ 
 = & \; {-}3.8978 \\ 
 {b\_2}/{\text{mol kg}^{-1}} = & \;0.02029 \\ 
 {b\_2} = & \;0.02029\;{\text{mol kg}^{-1}} \\ 
\end{aligned} $


Thus, taking into account the above-mentioned uncertainty, the calculated solubility of calcium hydroxide in water is


${b\_1} = \left( {{\text{0}}{\text{.02096}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T\_1} = 20\;^\circ\text{C}$


and


${b\_2} = \left( {{\text{0}}{\text{.02029}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T\_2} = 25\;^\circ\text{C}$.


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