Chapter 18, Problem 18.80P

Interpretation Introduction

Interpretation:

The ${\text{[H}}_{\text{2}}{\text{S], [HS}}^{\text{-}}{\text{], [S}}^{\text{2-}}{\text{], [H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{], pH, [OH}}^{\text{-}}\text{], and pOH}$ has to be calculated in $\text{0}\text{.10 M}$ solution of the diprotic hydro sulfuric acid.

Concept introduction:

An equilibrium constant $\left(K\right)$ is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general acid HA,

  $\text{HA}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_{\text{3}}{\text{O}}^{+}\left(\text{aq}\right)+{\text{A}}^{-}\left(\text{aq}\right)$

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  $K_a=\frac{\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}$                                                                                                   $\left(1\right)$

An equilibrium constant $\left(K\right)$ with subscript $a$ indicate that it is an equilibrium constant of an acid in water.

Explanation of Solution

Given,

Solution of $\text{0}\text{.1}\text{M}{\text{hydrosulfuric acid (H}}_{\text{2}}\text{S)}$.

Hydro sulfuric acid is diprotic acid therefore it can donate the two protons to the water.

The balance equation is given below,

  $\begin{array}{l}{\text{H}}_{\text{2}}\text{S (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ HS}}^{\text{-}}\text{(aq)}\\ \\ The value{\text{ K}}_{a1}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HS}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}\text{S}\right]}\\ \text{given,}\\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HS}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}\text{S}\right]}=\text{ 9}\text{.0}\times {10}^{-8}\end{array}$

The formed ${\text{HS}}^{\text{-}}$ ion can donate another proton, therefore the balance equation is given below,

$\begin{array}{l}{\text{HS}}^{\text{-}}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ S}}^{\text{2-}}\text{(aq)}\\ \\ The value{\text{ K}}_{a2}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{a2}\text{ =}\frac{\left[{\text{S}}^{\text{2-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HS}}^{\text{-}}\right]}\\ \text{given,}\\ {\text{K}}_{a2}\text{ =}\frac{\left[{\text{S}}^{\text{2-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HS}}^{\text{-}}\right]}=\text{ 1}\text{.0}\times {10}^{-17}\end{array}$

From the dissociation, ${\text{K}}_{a1}\rangle \rangle {\text{K}}_{a2}$, therefore the first dissociation forms complete ${\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}$ ion.

 $\text{0}{\text{.10 M H}}_{\text{2}}\text{S}$ Solution.

Therefore,

ICE table:

  ${\text{H}}_{\text{2}}\text{S (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ HS}}^{\text{-}}\text{(aq)}$

Initial concentration	0.1 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.1-x		x	x

The initial concentration is $\text{0}{\text{.10 M H}}_{\text{2}}\text{S}$.

  $\begin{array}{l}{\text{H}}_{\text{2}}\text{S (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ HS}}^{\text{-}}\text{(aq)}\\ \\ The value{\text{ K}}_{a1}iscalculatingbyu\sin gfollowingformula, \\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HS}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}\text{S}\right]}\\ \text{given,}\\ {\text{K}}_{a1}\text{ =}\frac{\left[{\text{HS}}^{\text{-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}\text{S}\right]}=\text{ 9}\text{.0}\times {10}^{-8}\end{array}$

Let consider,

  $\begin{array}{l}{\text{K}}_{a1}\text{ =}\frac{\left[{\text{HS}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{H}}_{\text{2}}\text{S}\right]}=\text{ 9}\text{.0}\times {10}^{-8}\\ \\ \frac{x^2}{\left(0.1-x\right)}=\text{9}\text{.0}\times {10}^{-8}\end{array}$

Assume, x is negligible so $0.1–x\approx 0.1$

  $\begin{array}{l}\frac{x^2}{\left(0.1-x\right)}=\text{9}\text{.0}\times {10}^{-8}\\ \\ x^2=9.0\times {10}^{-9}\\ \\ x=9.49\times {10}^{-5}\end{array}$

Hence, $\left[{\text{HS}}^{-}\right]=\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]=9.49\times {10}^{-5}$

Therefore,

  $\begin{array}{c}\text{pH}\text{=}\text{-}\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \\ \text{pH}\text{=}\text{-}\text{log}\left(9.49\times {10}^{-5}\right)\\ \\ \text{pH}\text{=}4.02\end{array}$

$\text{pH}$ of $\text{0}{\text{.1 M H}}_{\text{2}}\text{S}$ is $4.02$.

The $\left[{\text{OH}}^{\text{-}}\right]$ value is calculating by using following formula,

  $\begin{array}{l}\left[{\text{OH}}^{-}\right]\text{ = }\frac{{\text{K}}_{\text{w}}}{\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}\\ \\ \left[{\text{OH}}^{-}\right]\text{ = }\frac{{\text{1×10}}^{\text{-14}}}{9.49\times {10}^{-5}}\\ \\ \left[{\text{OH}}^{-}\right]\text{ = 1}{\text{.05×10}}^{\text{-10}}\end{array}$

$\left[{\text{OH}}^{-}\right]$ is $\text{1}{\text{.05×10}}^{\text{-10}}$

Therefore,

  $\begin{array}{c}\text{pOH}\text{=}\text{-}\text{log}\left[{\text{OH}}^{-}\right]\\ \\ \text{pOH}\text{=}\text{-}\text{log}\left(\text{1}{\text{.05×10}}^{\text{-10}}\right)\\ \\ \text{pOH}\text{=}10.0\end{array}$

pOH of the solution is $10.0$

The hydro sulfuric acid concentration at equilibrium is ${\left[{\text{H}}_{\text{2}}\text{S}\right]}_{\text{init}}\text{ – }{\left[{\text{H}}_{\text{2}}\text{S}\right]}_{\text{dissoc}}$

The hydro sulfuric acid concentration at equilibrium is $\text{0}\text{.1 – }9.49\times {10}^{-5}\text{M}$

The hydro sulfuric acid concentration at equilibrium is $\text{0}\text{.099}\text{M}$

Therefore, the ${\text{K}}_{a2}$ expression is given below for calculating ${\text{[S}}^{\text{2–}}\text{]}$, and $\left[{\text{HS}}^{\text{–}}\right]\text{ and }\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]$ come mostly from the first dissociation. This new calculation will have a new x value,

Therefore,

ICE table:

  ${\text{HS}}^{\text{-}}\text{ (aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\underset{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\text{(aq) }\text{+}{\text{ S}}^{\text{2-}}\text{(aq)}$

Initial concentration	$9.49\times {10}^{-5}$ M	-	$9.49\times {10}^{-5}$ M	0
Change	-x		+ x	+ x
At equilibrium	$9.49\times {10}^{-5}$ M - x		$9.49\times {10}^{-5}$ M + x	x

  $\begin{array}{c}The value{\text{ K}}_{a2}iscalculatingbyu\sin gfollowingformula, \\ \\ {\text{K}}_{a2}\text{ =}\frac{\left[{\text{S}}^{\text{2-}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}\right]}{\left[{\text{HS}}^{\text{-}}\right]}=\text{ 1}\text{.0}\times {10}^{-17}\\ \\ {\text{K}}_{a2}\text{=}\frac{\left(9.49\times {10}^{-5}+x\right)\times x}{\left(9.49\times {10}^{-5}-x\right)}=\text{1}\text{.0}\times {10}^{-17}\\ \\ {\text{K}}_{a2}\text{=}\frac{\left(9.49\times {10}^{-5}\right)\times x}{\left(9.49\times {10}^{-5}-x\right)}=\text{1}\text{.0}\times {10}^{-17}\\ \\ \left[{\text{S}}_{}^{\text{2-}}\right]\text{= 1}\text{.0}\times {10}^{-17}\text{M}\end{array}$

X value is very small, hence not necessary to calculate the ${\text{HS}}^{\text{-}}\text{ and}{\text{H}}_{\text{3}}{\text{O}}^{\text{+ }}$.

$\left[{\text{S}}_{}^{\text{2-}}\right]$ is $\text{1}\text{.0}\times {10}^{-17}\text{M}$