Ksp of calcium hydroxide [closed]

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?

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According to IUPAC Solubility Data Series Volume 52, Alkaline Earth Hydroxides in Water and Aqueous Solutions (1992), 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.

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 ${T_2} = 25\;^\circ\text{C}$.

• Thank you. Ah, so the $T/K$ value was just to get rid of the units. – Anonymous Mar 4 '15 at 2:25
• @Anonymous Yes. Generally, such numerical value equations are deprecated; the use of quantity equations is normally preferred and is strongly recommended. However, logarithms of units are ugly; thus, it is easier to eliminate the units and use a numerical value equation in this case. Anyway, this equation is not based on any chemical theory; it is just an arbitrary equation that sufficiently describes the experimental data. – Loong Mar 4 '15 at 13:07