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Mixing $\ce{Ca(OH)_2}$ with $\pu{0.1 M}$ concentration of $\ce{Ca(NO3)2}$.

What will actually happen in the process? What I will see with my eyes? I know that $\ce{Ca^2+}$ cation with concentration of $\pu{0.1 M}$ will be the first to react with $\ce{OH-}$, am I right? Would there be any changes in color? What will be the timing to the end point (approximately) of the reaction?

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Not much is going to happen. Due to low solubility of $\ce{Ca(OH)2}$, further suppressed by the presence of calcium cations from the dissolved $\ce{Ca(NO3)2}$, you get a suspension of calcium hydroxide, which unfiltered looks like white cloudy liquid and is also known as milk of lime.


Regarding the $\mathrm{pH}$ if the solution, $\ce{Ca(NO3)2}$ is prone to hydrolysis and alone would create slightly acidic medium:

Step 1:

$$ \begin{align} \ce{Ca(NO3)2 + H2O &<=> CaOHNO3 + HNO3} \tag{R1.1}\\ \ce{Ca^2+ + 2 NO3- + H2O &<=> CaOH+ + 2 NO3- + \color{red}{H+}}\tag{R1.2}\\ \ce{Ca^2+ + H2O &<=> CaOH+ + \color{red}{H+}}\tag{R1.3} \end{align} $$

Step 2:

$$ \begin{align} \ce{CaOHNO3 + H2O &<=> Ca(OH)2 + HNO3}\tag{R2.1}\\ \ce{CaOH+ + NO3- + H2O &<=> Ca(OH)2 + NO3- + \color{red}{H+}}\tag{R2.2}\\ \ce{CaOH+ + H2O &<=> Ca(OH)2 + \color{red}{H+}}\tag{R2.3} \end{align} $$

(Here .1, .2, .3 stand for molecular, complete ionic, and net ionic equation, correspondingly.)

But, when you add enough $\ce{Ca(OH)2}$ (saturated solution), which is a strong base:

$$\ce{Ca(OH)2 <=> Ca^2+ + 2 \color{blue}{OH-}}\, ,$$

I'd expect the resulting solution to be basic.

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