Chapter 15, Problem 84QRT

(a)

Interpretation Introduction

Interpretation:

The concentration of hydroxide ion has to be calculated at which magnesium hydroxide gets precipitated.

Explanation of Solution

Given,

The concentration of ${\text{Mg}}^{\text{2+}}$ is $\text{0}\text{.050 M}$. Salts gets dissociation in to their ions, ${\text{Mg(OH)}}_{\text{2}}$salt dissociate forms ${\text{Mg}}^{\text{2+}}$ ion and ${\text{2OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Mg(OH)}}_{\text{2}}$is given below,

    ${\text{Mg(OH)}}_{\text{2}}\underset{}{\rightleftharpoons }{\text{Mg}}^{\text{2+}}\text{+}{\text{2OH}}^{\text{-}}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is to be calculated for ${\text{Mg(OH)}}_{\text{2}}$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Mg}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Mg}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\text{=}\text{1}{\text{.8×10}}^{\text{-11}}\\ {\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\text{=}\frac{\text{1}{\text{.8×10}}^{\text{-11}}}{\left[{\text{Mg}}^{\text{2+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\sqrt{\frac{\text{1}{\text{.8×10}}^{\text{-11}}}{\text{0}\text{.050}\text{M}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\sqrt{\text{3}{\text{.6×10}}^{\text{-10}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{1}{\text{.897×10}}^{\text{-5}}M\end{array}$

Therefore, ${\text{Mg(OH)}}_{\text{2}}$starts to precipitated when the concentration of hydroxide ion is $\text{1}{\text{.897×10}}^{\text{-5}}\text{M}$.

(b)

Interpretation Introduction

Interpretation:

At concentration of hydroxide ion, other cation will get precipitated or not has to be determined.

Explanation of Solution

Given,

1. (1) The concentration of ${\text{Na}}^{\text{+}}$ is $\text{0}\text{.46 M}$. Salts gets dissociation in to their ions, $\text{NaOH}$salt dissociate forms ${\text{Na}}^{\text{+}}$ ion and ${\text{OH}}^{\text{-}}$ ions.

The dissociation of $\text{NaOH}$is given below,

    $\text{NaOH}\underset{}{\rightleftharpoons }{\text{Na}}^{\text{+}}\text{+}{\text{OH}}^{\text{-}}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is so large for $\text{NaOH}$.

Therefore, when ${\text{K}}_{\text{sp}}$ is large no precipitation will occur.

Given,

1. (2) The concentration of ${\text{Ca}}^{\text{2+}}$ is $\text{0}\text{.01 M}$

Salts gets dissociation in to their ions, ${\text{Ca(OH)}}_{\text{2}}$salt dissociate forms ${\text{Ca}}^{\text{2+}}$ ion and ${\text{2OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Ca(OH)}}_{\text{2}}$is given below,

    ${\text{Ca(OH)}}_{\text{2}}\underset{}{\rightleftharpoons }{\text{Ca}}^{\text{2+}}\text{+}{\text{2OH}}^{\text{-}}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is to be calculated for ${\text{Ca(OH)}}_{\text{2}}$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ca}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Ca}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\text{=}\text{5}{\text{.5×10}}^{\text{-6}}\\ \text{Where,}\\ \text{Q}\text{=}{\left[{\text{Ca}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\\ \text{Q}\text{=}\left(\text{0}\text{.01M}\right){\left(\text{1}{\text{.9×10}}^{\text{-5}}\text{M}\right)}^{\text{2}}\\ \text{Q}\text{=3}{\text{.61×10}}^{\text{-12}}\text{M}\end{array}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is large for ${\text{Ca(OH)}}_{\text{2}}$.

Therefore, when ${\text{K}}_{\text{sp}}$ is large no precipitation will occur.

1. (3) The concentration of ${\text{Al}}^{\text{3+}}$ is ${\text{4×10}}^{\text{-7}}\text{ M}$

Salts gets dissociation in to their ions, ${\text{Al(OH)}}_3$salt dissociate forms ${\text{Al}}^{\text{3+}}$ ion and ${\text{3OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Al(OH)}}_3$is given below,

    ${\text{Al(OH)}}_3\underset{}{\rightleftharpoons }{\text{Al}}^{\text{3+}}\text{+}{\text{3OH}}^{\text{-}}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is to be calculated for ${\text{Al(OH)}}_3$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\text{=}\text{1}{\text{.3×10}}^{\text{-33}}\\ \text{Where,}\\ \text{Q}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\\ \text{Q}\text{=}\left({\text{4×10}}^{\text{-7}}\text{ M}\right){\left(\text{1}{\text{.9×10}}^{\text{-5}}\text{M}\right)}^3\\ \text{Q}\text{=2}{\text{.74×10}}^{\text{-21}}\text{M}\end{array}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is smaller for ${\text{Al(OH)}}_3$.

Therefore, when ${\text{K}}_{\text{sp}}$ is small, precipitation will occur. Hence, ${\text{Al(OH)}}_3$starts to precipitated.

1. (4) The concentration of ${\text{Fe}}^{\text{2+}}$ is ${\text{2×10}}^{\text{-7}}\text{ M}$

Salts gets dissociation in to their ions, ${\text{Fe(OH)}}_2$salt dissociate forms ${\text{Fe}}^{\text{2+}}$ ion and ${\text{OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Fe(OH)}}_2$is given below,

    ${\text{Fe(OH)}}_2\underset{}{\rightleftharpoons }{\text{Fe}}^{\text{2+}}\text{+}2{\text{OH}}^{\text{-}}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is to be calculated for ${\text{Fe(OH)}}_2$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Fe}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^2\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Fe}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^2\text{=}\text{8}{\text{.0×10}}^{\text{-16}}\\ \text{Where,}\\ \text{Q}\text{=}{\left[{\text{Fe}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^2\\ \text{Q}\text{=}\left({\text{2×10}}^{\text{-7}}\text{ M}\right){\left(\text{1}{\text{.9×10}}^{\text{-5}}\text{M}\right)}^2\\ \text{Q}\text{=7}{\text{.22×10}}^{\text{-17}}\text{M}\end{array}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is large for ${\text{Fe(OH)}}_2$.

Therefore, when ${\text{K}}_{\text{sp}}$ is large no precipitation will occur. In conclusion, the Solubility product constant (${\text{K}}_{\text{sp}}$) is smaller for ${\text{Al(OH)}}_3$. Precipitation will occur only for ${\text{Al(OH)}}_3$.

(c)

Interpretation Introduction

Interpretation:

The percent has to be calculated when the other cation gets precipitated.

Explanation of Solution

Given,

1. (a) The concentration of ${\text{Mg}}^{\text{2+}}$ is $\text{0}\text{.050 M}$.  The enough hydroxide is added to precipitate 50% of the ${\text{Mg}}^{\text{2+}}$.

Salts gets dissociation in to their ions, ${\text{Mg(OH)}}_{\text{2}}$salt dissociate forms ${\text{Mg}}^{\text{2+}}$ ion and ${\text{2OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Mg(OH)}}_{\text{2}}$is given below,

    ${\text{Mg(OH)}}_{\text{2}}\underset{}{\rightleftharpoons }{\text{Mg}}^{\text{2+}}\text{+}{\text{2OH}}^{\text{-}}$

Solubility product constant (Ksp) is to be calculated for ${\text{Mg(OH)}}_{\text{2}}$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

Therefore, the concentration of ${\text{Mg}}^{\text{2+}}$ is calculated as follows,

  $\begin{array}{l}{\text{Mg}}^{\text{2+}}\text{=}\text{(0}\text{.50M)}\times \text{(0}\text{.050M)}\\ {\text{Mg}}^{\text{2+}}\text{=}\text{0}\text{.025}\text{M}\end{array}$

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Mg}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Mg}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\text{=}\text{1}{\text{.8×10}}^{\text{-11}}\\ {\left[{\text{OH}}^{\text{-}}\right]}^{\text{2}}\text{=}\frac{\text{1}{\text{.8×10}}^{\text{-11}}}{\left[{\text{Mg}}^{\text{2+}}\right]}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\sqrt{\frac{\text{1}{\text{.8×10}}^{\text{-11}}}{\text{0}\text{.025}\text{M}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\sqrt{\text{7}{\text{.2×10}}^{\text{-10}}}\\ \left[{\text{OH}}^{\text{-}}\right]\text{=}\text{2}{\text{.68×10}}^{\text{-5}}\text{M}\end{array}$

Therefore, ${\text{Mg(OH)}}_{\text{2}}$starts to precipitated when the concentration of hydroxide ion is $\text{2}{\text{.68×10}}^{\text{-5}}\text{M}$.

1. (b) Solubility product constant (Ksp) is so large for $\text{NaOH}$. Therefore, when the Ksp is large no precipitation will occur. Sodium ion does not for precipitated with $\text{NaOH}$, therefore, ${\text{Na}}^{\text{+}}$ ion is zero percentage precipitated.
2. (c) Solubility product constant (Ksp) is large for ${\text{Ca(OH)}}_{\text{2}}$. Therefore, when the Ksp is large no precipitation will occur.
3. (d) The concentration of ${\text{Al}}^{\text{3+}}$ is ${\text{4×10}}^{\text{-7}}\text{ M}$. Salts gets dissociation in to their ions, ${\text{Al(OH)}}_3$salt dissociate forms ${\text{Al}}^{\text{3+}}$ ion and ${\text{3OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Al(OH)}}_3$is given below,

    ${\text{Al(OH)}}_3\underset{}{\rightleftharpoons }{\text{Al}}^{\text{3+}}\text{+}{\text{3OH}}^{\text{-}}$

Solubility product constant (Ksp) is to be calculated for ${\text{Al(OH)}}_3$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\text{=}\text{1}{\text{.3×10}}^{\text{-33}}\\ \\ \text{Where,}\\ \\ \text{Q}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\\ \text{Q}\text{=}\left({\text{4×10}}^{\text{-7}}\text{ M}\right){\left(\text{2}{\text{.7×10}}^{\text{-5}}\text{M}\right)}^3\\ \text{Q}\text{=7}{\text{.7×10}}^{\text{-21}}\text{M}\end{array}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is smaller for ${\text{Al(OH)}}_3$. Therefore, when the Ksp is small, precipitation will occur. Hence, ${\text{Al(OH)}}_3$starts to precipitated.

The ICE table is given below,

  ${\text{Al}}^{\text{3+}}(aq)\underset{}{\rightleftharpoons }{\text{3OH}}^{\text{-}}\text{(aq)}\text{+}{\text{Al(OH)}}_{\text{3}}$

Initial concentration	${\text{4×10}}^{\text{-7}}\text{ M}$	$\text{2}{\text{.7×10}}^{\text{-5}}\text{M}$	0
Change	-${\text{4×10}}^{\text{-7}}\text{ M}$	-3(${\text{4×10}}^{\text{-7}}\text{ M}$)	solid
Final conc.	0	$\text{2}{\text{.6×10}}^{\text{-5}}\text{M}$	solid

At equilibrium: ${\text{Al(OH)}}_3\underset{}{\rightleftharpoons }{\text{Al}}^{\text{3+}}\text{+}{\text{3OH}}^{\text{-}}$

The ICE table is given below,

  ${\text{Al(OH)}}_3\underset{}{\rightleftharpoons }{\text{Al}}^{\text{3+}}(aq)\text{+}{\text{3OH}}^{\text{-}}(aq)$

Initial concentration	solid	0	$\text{2}{\text{.6×10}}^{\text{-5}}\text{M}$
Change	-x	+x	+x
At equilibrium.	solid	+x	$\text{2}{\text{.6×10}}^{\text{-5}}\text{M}\text{+}\text{x}$

At equilibrium: ${\text{Al(OH)}}_3\underset{}{\rightleftharpoons }{\text{Al}}^{\text{3+}}\text{+}{\text{3OH}}^{\text{-}}$

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\text{=}\text{1}{\text{.3×10}}^{\text{-33}}\end{array}$

Where,

  $\begin{array}{l}{\left[{\text{Al}}^{\text{3+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^{\text{3}}\text{=}\text{1}{\text{.3×10}}^{\text{-33}}\\ \left(\text{x}\right){\left(\text{2}{\text{.6×10}}^{\text{-5}}\text{M+x}\right)}^{\text{3}}\text{=}\text{1}{\text{.3×10}}^{\text{-33}}\end{array}$

Let consider x is small,

  $\begin{array}{l}\left(\text{x}\right){\left(\text{2}{\text{.6×10}}^{\text{-5}}\text{M}\right)}^{\text{3}}\text{=}\text{1}{\text{.3×10}}^{\text{-33}}\\ \left(\text{x}\right)\text{=}\frac{\text{1}{\text{.3×10}}^{\text{-33}}}{{\left(\text{2}{\text{.6×10}}^{\text{-5}}\text{M}\right)}^{\text{3}}}\\ \left(\text{x}\right)\text{=}\left[{\text{Al}}^{\text{3+}}\right]\text{=}\text{7}{\text{.4×10}}^{\text{-20}}\end{array}$

The concentration of ${\text{Al}}^{\text{3+}}$ is given below,

  $\begin{array}{l}\text{Concentartion of }{\text{Al}}^{\text{3+}}\text{precipitated = Initial}{\text{concentartion of Al}}^{\text{3+}}\text{-}\left[{\text{Al}}^{\text{3+}}\right]\\ {\text{Concentartion of Al}}^{\text{3+}}\text{precipitated = 4}{\text{.0×10}}^{\text{-7}}\text{-7}{\text{.4×10}}^{\text{-20}}\text{M}\\ {\text{Concentartion of Al}}^{\text{3+}}\text{precipitated = 4}{\text{.0×10}}^{\text{-7}}\text{M}\end{array}$

Therefore,

  $\begin{array}{l}\text{Percent dissociated = }\frac{\text{dissociation}}{\text{initial}}\text{×100}\\ \text{Percent dissociated = }\frac{\text{ 4}{\text{.0×10}}^{\text{-7}}\text{M}}{\text{ 4}{\text{.0×10}}^{\text{-7}}\text{M}}\times 100\%\\ \text{Percent dissociated = }100\%\end{array}$

The percentage of ${\text{Al(OH)}}_3$ is $100\%$.

1. (e) The concentration of ${\text{Fe}}^{\text{2+}}$ is ${\text{2×10}}^{\text{-7}}\text{ M}$ Salts gets dissociation in to their ions, ${\text{Fe(OH)}}_2$salt dissociate forms ${\text{Fe}}^{\text{2+}}$ ion and ${\text{OH}}^{\text{-}}$ ions.

The dissociation of ${\text{Fe(OH)}}_2$is given below,

    ${\text{Fe(OH)}}_2\underset{}{\rightleftharpoons }{\text{Fe}}^{\text{2+}}\text{+}2{\text{OH}}^{\text{-}}$

Solubility product constant (Ksp) is to be calculated for ${\text{Fe(OH)}}_2$.

The ${\text{K}}_{\text{sp}}$ value is calculated as follows,

  $\begin{array}{l}{\text{K}}_{\text{sp}}\text{=}{\left[{\text{Fe}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^2\\ {\text{K}}_{\text{sp}}\text{=}{\left[{\text{Fe}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^2\text{=}\text{8}{\text{.0×10}}^{\text{-16}}\end{array}$

Where,

  $\begin{array}{l}\text{Q}\text{=}{\left[{\text{Fe}}^{\text{2+}}\right]}^{\text{1}}{\left[{\text{OH}}^{\text{-}}\right]}^2\\ \text{Q}\text{=}\left({\text{2×10}}^{\text{-7}}\text{ M}\right){\left(\text{2}{\text{.7×10}}^{\text{-5}}\text{M}\right)}^2\\ \text{Q}\text{=1}{\text{.4×10}}^{\text{-116}}\text{M}\end{array}$

Solubility product constant (${\text{K}}_{\text{sp}}$) is large for ${\text{Fe(OH)}}_2$.  Therefore, when ${\text{K}}_{\text{sp}}$ is large no precipitation will occur.

In conclusion, Precipitation will occur only for ${\text{Al(OH)}}_3$.${\text{Al(OH)}}_3$ shows $100\%$ precipitation.