Question

The first ionization constant of $H_{2}S$ is 9.1×$10^{-8}$. Calculate the concentration of $HS^{-}$ ions in its 0. 1 M solution. How will this concentration be affected if the solution is 0.1 M in HCl also ? If the second dissociation constant of $H_{2}S$ is 1.2 × $10^{-13}$, calculate the concentration of $S^{2-}$ under both conditions.

Answer 1

HEY MATE YOUR ANSWER IS

Intermolecular forces. Silicon dioxide is really strange with its shape. You'd assume it to be linear like carbon dioxide but it holds a more tetrahedral orientation. This polarizes the molecule and helps them form complex chains of silicon dioxide. It's hard to find black and white information on this as many articles will contradict each other. But we do use silica as a polar surface for TLC and glass does have a slight attraction to water.

So I guess that's the answer. Silicon dioxide is polar. Carbon dioxide is not.

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Answer 2

"(i) To calculate the concentration of ${ HS }^{ - }$ ion (absence of HCl)

Let's write the equilibrium chemical reaction

$\begin{matrix} \quad  \\ { C }_{ initial } \\ { C }_{ final } \end{matrix}\begin{matrix} { H }_{ 2 }S \\ 0.1 \\ 0.1-x \end{matrix}\begin{matrix} { \leftrightarrow \quad H }^{ + }\quad + \\ 0 \\ x \end{matrix}\begin{matrix} { HS }^{ - } \\ 0 \\ x \end{matrix}$

${ K }_{ a1 }\quad =\quad \frac { [{ H }^{ + }][{ HS }^{ - }] }{ { [H }_{ 2 }S] }$

$9.1\quad \times \quad { 10 }^{ -8 }\quad =\quad \frac { x.x }{ 0.1\quad -\quad x }$

$(9.1\quad \times \quad { 10 }^{ -8 })\quad (0.1\quad -\quad x)\quad =\quad { x }^{ 2 }$

0.1 - x is considered as 0.1 M

$(9.1\quad \times \quad { 10 }^{ -8 })\quad (0.1)\quad =\quad { x }^{ 2 }$

$x\quad =\quad \sqrt { 9.1\quad \times \quad { 10 }^{ -8 } }$

$x\quad =\quad 9.54\quad \times \quad { 10 }^{ -5 }\quad M$

Therefore, $[H{ S }^{ - }]\quad =\quad 9.54\quad \times \quad { 10 }^{ -5 }\quad M$

(i) To calculate the concentration of ${ HS }^{ - }$ ion (presence of HCl)

Let's write the equilibrium chemical reaction

$\begin{matrix} \quad  \\ { C }_{ initial } \\ { C }_{ final } \end{matrix}\begin{matrix} { H }_{ 2 }S \\ 0.1 \\ 0.1-y \end{matrix}\begin{matrix} { \leftrightarrow \quad H }^{ + } \\ \quad \quad 0 \\ \quad \quad y \end{matrix}\begin{matrix} +\quad { HS }^{ - } \\ 0 \\ y \end{matrix}$

$\begin{matrix} HCl\quad \leftrightarrow \quad  \end{matrix}\begin{matrix} { H }^{ + } \\ 0.1 \end{matrix}\begin{matrix} { \quad +\quad Cl }^{ - } \\ \quad \quad 0.1 \end{matrix}$

${ K }_{ a1 }\quad \quad =\quad \frac { [{ H }^{ + }][{ HS }^{ - }] }{ { [H }_{ 2 }S] }$

${ K }_{ a1 }=\quad \frac { [y][0.1\quad +\quad y] }{ [0.1\quad -\quad y] }$

$9.1\quad \times \quad { 10 }^{ -8 }\quad =\quad y\quad \times \quad \frac { [0.1\quad +\quad y] }{ [0.1\quad -\quad y] }$

0.1 - y is considered as 0.1 M

$(9.1\quad \times \quad { 10 }^{ -8 })(0.1)\quad =\quad { x }^{ 2 }$

$9.1\quad \times \quad { 10 }^{ -8 }\quad =\quad y$

$[H{ S }^{ - }]\quad =\quad 9.54\quad \times \quad { 10 }^{ -8 }$

(ii) Calculate the concentration of ${ S }^{ 2- }$ (in the absence of 0.1 M of HCl)

${ HS }^{ - }\quad \leftrightarrow \quad { H }^{ +\quad  }+\quad { S }^{ 2- }$

$H{ S }^{ - }\quad =\quad 9.54\quad \times \quad { 10 }^{ -5 }\quad M$

${ S }^{ 2- }\quad =\quad x$

Also, $[{ H }^{ + }]\quad =\quad 9.54\quad \times \quad { 10 }^{ -5 }\quad M$

${ K }_{ a2 }\quad =\quad \frac { (9.54\quad \times \quad { 10 }^{ -5 })(x) }{ (9.54\quad \times \quad { 10 }^{ -5 }) } \quad =\quad 1.2\quad \times \quad { 10 }^{ -5 }\quad M$

$x\quad =\quad { S }^{ 2- }\quad =\quad 1.2\quad \times \quad { 10 }^{ -5 }\quad M$

(ii) Calculate the concentration of ${ S }^{ 2- }$ (in the presence of 0.1 M of HCl)

${ HS }^{ - }\quad \leftrightarrow \quad { H }^{ +\quad  }+\quad { S }^{ 2- }$

$H{ S }^{ - }\quad =\quad 9.1\quad \times \quad { 10 }^{ -5 }\quad M$

$[{ H }^{ + }]\quad =\quad 0.1\quad M$

$[{ S }^{ 2- }]\quad =\quad { x }^{ 1 }\\ { K }_{ a2 }\quad =\quad \frac { [{ H }^{ + }][{ S }^{ 2- }] }{ [H{ S }^{ - }] }$

$1.2\quad \times \quad { 10 }^{ -13\quad  }=\quad \frac { (0.1)({ x }^{ 1 }) }{ (9.1\quad \times \quad { 10 }^{ -8 }) }$

$10.92\quad \times \quad { 10 }^{ -21 }\quad =\quad 0.1{ x }^{ 1 }$

${ x }^{ 1 }\quad =\quad \frac { 10.92\quad \times \quad { 10 }^{ -21 } }{ 0.1 }$

${ x }^{ 1 }\quad =\quad 1.092\quad \times \quad { 10 }^{ -19 }\quad M$

${ K }_{ a1 }\quad =\quad 1.74\quad \times \quad { 10 }^{ -5 }\quad M$"