Question

Calculate the solubility of mn(oh)2 in grams per liter when buffered at ph=7.0. assume that buffer capacity is not exhausted

Answer 1

The solubility of ${\text{Mn}}{\left( {{\text{OH}}} \right)_2}$ in grams per litre is $\boxed{0.002001{\text{ g/L}}}$.

Further explanation:

Solubility product constant:

It is the equilibrium constant for the reaction that comes into play when an ionic compound is dissolved to produce its respective ions. It is represented by ${{\text{K}}_{{\text{sp}}}}$. Consider ${{\text{A}}_x}{{\text{B}}_y}$ to be an ionic compound. Its dissociation occurs as follows:

 ${{\text{A}}_x}{{\text{B}}_y} \rightleftharpoons x{{\text{A}}^{y + }} + y{{\text{A}}^{x - }}$

The expression for its ${{\text{K}}_{{\text{sp}}}}$ is as follows:

 ${{\text{K}}_{{\text{sp}}}} = {\left[ {{{\text{A}}^{y + }}} \right]^x}{\left[ {{{\text{B}}^{x - }}} \right]^y}$

Here,

${{\text{K}}_{{\text{sp}}}}$ is the solubility product constant.

$\left[ {{{\text{A}}^ + }} \right]$ is the concentration of  ions.

$\left[ {{{\text{B}}^ - }} \right]$ is the concentration of  ions.

The relation between pH and pOH of the solution is given by the following formula:

${\text{pH}} + {\text{pOH}} = {\text{14}}$   ...... (1)                                                                

Rearrange equation (1) to calculate the pOH of the solution.

${\text{pOH}} = {\text{14}} - {\text{pH}}$  ...... (2)                                                                  

The value of pH is 7. Substitute this value in equation (2).

 $\begin{aligned}{\text{pOH}} &= {\text{14}} - {\text{7}}\\&= {\text{7}}\\\end{aligned}$

The expression for pOH is as follows:

${\text{pOH}} = - \log \left[ {{\text{O}}{{\text{H}}^ - }} \right]$  ...... (3)                                                              

Rearrange equation (3) for $\left[ {{\text{O}}{{\text{H}}^ - }} \right]$.

 $\left[ {{\text{O}}{{\text{H}}^ - }} \right] = {10^{ - {\text{pOH}}}}$ ...... (4)                                                                

The value of pOH is 7. Substitute this value in equation (4).

 $\left[ {{\text{O}}{{\text{H}}^ - }} \right] = {10^{ - 7}}{\text{ M}}$

The dissociation reaction of ${\text{Mn}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ is as follows:

 ${\text{Mn}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {aq} \right) \rightleftharpoons {\text{M}}{{\text{n}}^{2 + }}\left( {aq} \right) + 2{\text{O}}{{\text{H}}^ - }\left( {aq} \right)$

Let us consider the solubility of ${\text{Mn}}{\left( {{\text{OH}}} \right)_2}$ be s. Therefore, after dissociation, the concentration of ${\text{M}}{{\text{n}}^{2 + }}$ and ${\text{O}}{{\text{H}}^ - }$ are s and ${10^{ - 7}} + 2s$ respectively.

The formula to calculate the molar solubility of  is as follows:

${{\text{K}}_{{\text{sp}}}} = \left[ {{\text{M}}{{\text{n}}^{{\text{2 + }}}}}\right]{\left[ {{\text{O}}{{\text{H}}^ - }}\right]^2}$  …… (5)                                                                                        

Substitute s for $\left[ {{\text{M}}{{\text{n}}^{{\text{2 + }}}}} \right]$, ${10^{ - 7}} + 2s$ for $\left[ {{\text{O}}{{\text{H}}^ - }} \right]$ and ${\text{4}}{\text{.6}} \times {10^{ - 14}}$ for ${{\text{K}}_{{\text{sp}}}}$ in equation (5).

${\text{4}}{\text{.6}} \times {10^{ - 14}} = \left( s \right){\left( {{{10}^{ - 7}} + 2s} \right)^2}$  

Solve for s,

$s = 2.25 \times {10^{ - 5}}{\text{ M}}$

Therefore the solubility of ${\text{Mn}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ comes out to be $2.25 \times {10^{ - 5}}{\text{ M}}$.

The formula to calculate the solubility of ${\text{Mn}}{\left( {{\text{OH}}} \right)_{\text{2}}}$ in g/L is as follows:

$\begin{aligned}{\text{Solubility of Mn}}{\left( {{\text{OH}}} \right)_{\text{2}}}\left( {{\text{g/L}}} \right) &= \left( {{\text{Solubility of Mn}{{\left({{\text{OH}}}\right)}_{\text{2}}}\left( {{\text{mol/L}}} \right)} \right)\\\left( {{\text{Molar mass of Mn}}{{\left( {{\text{OH}}} \right)}_{\text{2}}}} \right) \\ \end{aligned}$                …… (6)

Substitute $2.25 \times {10^{ - 5}}{\text{ M}}$ for solubility and 88.95 g/mol in equation (6).

 $\begin{aligned}{\text{Solubility of Mn}}{\left( {{\text{OH}}} \right)_{\text{2}}} &= \left( {\frac{{2.25 \times {{10}^{ - 5}}{\text{ mol}}}}{{{\text{1 L}}}}} \right)\left({\frac{{{\text{88}}{\text{.95 g}}}}{{{\text{1 mol}}}}} \right)\\&= 0.002001{\text{ g/L}}\\\end{aligned}$

Learn more:

1. Sort the solubility of gas will increase or decrease: brainly.com/question/2802008.
2. What is the pressure of the gas? brainly.com/question/6340739.

Answer details:

Grade: School School

Subject: Chemistry

Chapter: Chemical equilibrium

Keywords: solubility, Mn2+, 2OH-, Mn(OH)2, Ksp, solubility product, molar mass, 88.95 g/mol, 0.002001 g/L.

Answer 2

When PH + POH = 14 
∴ POH = 14 -7 = 7

when POH = -㏒[OH-]

          7    = -㏒ [OH-]
∴[OH-] = 10^-7

by using ICE table:

           Mn(OH)2(s) ⇄  Mn2+ (aq)  + 2OH-(aq)
initial                              0                     10^-7
change                           +X                      +2X
Equ                                 X                  (10^-7 + 2X)

when Ksp = [Mn2+][OH-]^2

when Ksp of Mn(OH)2 = 4.6 x 10^-14

by substitution:

4.6 x 10^-14 = X*(10^-7+2X)^2  by solving this equation for X

∴ X =2.3 x 10-5 M

∴ The solubility of Mn(OH)2 in grams per liter (when the molar mass of Mn(OH)2 = 88.953 g/mol
= 2.3 x10^-5 moles/L * 88.953 g/mol

= 0.002 g/ L