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Serhud [2]
3 years ago
11

The hydroxide ion concentration of a saturated solution of ca(oh)2 is m. what is the solubility product constant for ca(oh)2

Chemistry
1 answer:
dalvyx [7]3 years ago
6 0
Solubility product constants are values to describe the saturation of ionic compounds with low solubility. A saturated solution is when there is a dynamic equilibrium between the solute dissolved, the dissociated ions, the undissolved and the compound. It is calculated from the product of the ion concentration in the solution. For the base, Ca(OH)2, the dissociation would  be as follows:<span>

Ca(OH)2 = Ca2+ + 2OH-

So, the expression for the solubility product constant would be as follows:

Ksp = [Ca2+] [OH-]^2

let x be the concentration of the Ca2+. So,

</span>
Ksp = [x] [2x]^2
<span>Ksp = 4x^3

You have to substitute the value of the concentration of the calcium hydroxide in the final expression which is not given in the problem statement in order to evaluate Ksp.
</span><span>
</span>
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Explanation:

The given data is as follows.

           n_{1} = 9,          n_{2} = ?

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As we know that,

                       Energy (E) = \frac{hc}{\lambda}

where,   h = planck's constant = 6.626 \times 10^{-34} Js

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According to Reydberg's equation, we will calculate the energy emitted by the photon as follows.

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or,     \frac{hc}{\lambda} = -2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]

Putting the given values into the above equation as follows.

         \frac{hc}{\lambda} = -2.179 \times 10^{-18} J \times (Z)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{n^{2}_{1}}]

        \frac{6.626 \times 10^{-34} Js \times 3 \times 10^{8}m/s}{384 \times 10^{-9} m} = -2.179 \times 10^{-18} J \times (1)^{2}[\frac{1}{n^{2}_{2}} - \frac{1}{(9)^{2}}]  

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lorasvet [3.4K]

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