# 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?

## closed as off-topic by Todd Minehardt, Mithoron, airhuff, A.K., Karsten TheisApr 8 at 2:46

This question appears to be off-topic. The users who voted to close gave this specific reason:

If this question can be reworded to fit the rules in the help center, please edit the question.

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

${T_1} = 20\;^\circ\text{C} = 293.15\;{\text{K}}$

\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

${T_2} = 25\;^\circ\text{C} = 298.15\;{\text{K}}$

\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}
${b_1} = \left( {{\text{0}}{\text{.02096}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ at a temperature of ${T_1} = 20\;^\circ\text{C}$
${b_2} = \left( {{\text{0}}{\text{.02029}} \pm 0.00017} \right)\;{\text{mol kg}^{-1}}$ 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