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?


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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