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Question: How would you calculate the molar solubility of \[Mn{\left( {OH} \right)_2}\] \[\left( {Ksp{\text{ }...

How would you calculate the molar solubility of Mn(OH)2Mn{\left( {OH} \right)_2} (Ksp = 1.6e−13)\left( {Ksp{\text{ }} = {\text{ }}1.6e - 13} \right) in pure water?

Explanation

Solution

Solubility is the result of the ionic fixations or exercises of an electrolyte and the molar solubility is the quantity of moles of salts that can be disintegrated in one liter of answer for structure an immersed arrangement and afterward determine the necessary condition which offers the response.

Complete step by step answer: In the classes of actual science, we have learned about the idea of molar solubility and furthermore about the solubility item and a portion of the overall definitions identifying with it.
Let us presently perceive how the solubility item can be determined from the molar solubility.
-Firstly, let us comprehend what solubility item is and what the molar solubility states.
-Solubility item is the greatest result of the ionic focuses or the exercises of an electrolyte that at one temperature will have the option to proceed in harmony with undissolved stage or in basic word say can say that Ksp{K_{sp}} is balance steady for a strong substance dissolving in fluid arrangement. Molar solubility is characterized as the quantity of moles of salts that can be broken up in one liter of answer for structure an immersed arrangement.
Let us currently accept the overall separation equilibrium as:
XnYm=n.Xm++m.Yn−{X_n}{Y_m} = n.{X^{m + }} + m.{Y^{n - }}
We will accept that we have molar solubility as ssmol L−1{L^{ - 1}} for this specific salt in water at room temperature and this portrays that we can break down ′s′'s' moles ofXnYm{X_n}{Y_m} per liter of arrangement at this specific temperature.
Here, each mole of XnYm{X_n}{Y_m}that breaks down will create nn moles ofXm+{X^{m + }}that are cations and mm moles of Yn−{Y^{n - }} that are anions.
This is only the soaked arrangement structures will have [Xm+]=n.s\left[ {{X^{m + }}} \right] = n.sand [Yn−]=m.s\left[ {{Y^{n - }}} \right] = m.s
Along these lines, the solubility for this harmony separation will be:
Ksp=(n×s)n×(m×s)m{K_{sp}} = {\left( {n \times s} \right)^n} \times {\left( {m \times s} \right)^m}
⇒Ksp=nn×sn×mm×sm\Rightarrow {K_{sp}} = {n^n} \times {s^n} \times {m^m} \times {s^m}
Ksp=nn×mm×s(n+m){K_{sp}} = {n^n} \times {m^m} \times {s^{\left( {n + m} \right)}}
So now start the estimation of the given question,

Ksp=1.6×10−13{K_{sp}} = 1.6 \times {10^{ - 13}}
Ksp=1.6×10−13=[Mn2+][OH−]2{K_{sp}} = 1.6 \times {10^{ - 13}} = \left[ {M{n^{2 + }}} \right]{\left[ {O{H^ - }} \right]^2}
On the off chance that we call the solubility of manganese (II)\left( {II} \right) hydroxide SS, at that point by the stoichiometry of the response:
Ksp=S(2S)2=4S3{K_{sp}} = S{\left( {2S} \right)^2} = 4{S^3}.
So, S=1.6×10−1343S = \sqrt[3]{{\frac{{1.6 \times {{10}^{ - 13}}}}{4}}}

Note:
Note that the molar solubility is identified with the solubility item consistent Ksp{K_{sp}} as the higher the estimation of the solubility item, more dissolvable the compound will be and this says that they are straightforwardly identified with one another.