Question

Assuming each solution to be 0.10 M , rank the following aqueous solutions in order of decreasing pH.

Rank the solutions from the highest to lowest pH. To rank items as equivalent, overlap them

N2H2

Ba(OH)2

HOCL

NAOH

HCL

Answer 1

Concepts and reason

The concept used is to arrange the aqueous solution of the given salts in the order of decreasing ${\rm{pH}}$ .

Fundamentals

$pH$ is the measure of hydrogen ion concentration. Lower the $pH$ , higher is the hydrogen ion concentration and lower is the hydroxide ion concentration.

Higher the $pH$ , lower is the hydrogen ion concentration and higher is the hydroxide ion concentration.

The hydrogen ion concentration is calculated as follows:

$\left[ {{{\rm{H}}^ + }} \right] = {10^{ - pH}}$

The hydroxide ion concentration $pOH$ and $pH$ are related as follows:

$\left[ {{{\rm{H}}^ + }} \right]\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 1.0 \times {10^{ - 14}}$

If the ${\rm{pH}} < 7$ , the solution is acidic.

If the ${\rm{pH}} > 7$ , the solution is basic.

If the ${\rm{pH}} = 7$ , the solution is neutral.

The dissociation of acid produces hydronium ions. The reaction is as follows:

${\rm{A}} + {{\rm{H}}_2}{\rm{O}}\longrightarrow{{}}{{\rm{A}}^ - }\left( {aq} \right) + {{\rm{H}}_3}{{\rm{O}}^ + }\left( {aq} \right)$

The dissociation of base produces hydroxide ions. The reaction is as follows:

${\rm{B}} + {{\rm{H}}_2}{\rm{O}}\longrightarrow{{}}{\rm{B}}{{\rm{H}}^ + }\left( {aq} \right) + {\rm{O}}{{\rm{H}}^ - }\left( {aq} \right)$

Among the given aqueous solutions, ${\rm{HCl}}$ and ${\rm{HOCl}}$ are the acids.

${\rm{HCl}}$ is a strong acid.

$\left[ {{\rm{HCl}}} \right] = 0.1{\rm{ M}}$

On dissociation, it produces one mole of ${{\rm{H}}^ + }$ .

So,

$\begin{array}{l}\\\left[ {{{\rm{H}}^ + }} \right] = 1 \times 0.1{\rm{ M}}\\\\{\rm{ = 0}}{\rm{.1 M}}\\\end{array}$

Since, the hydronium concentration is high, its ${\rm{pH}}$ is small.

${\rm{HOCl}}$ is a weak acid.

Its dissociation is not complete.

So,

$\left[ {{{\rm{H}}^ + }} \right] < {\rm{0}}{\rm{.1 M}}$

Since, the concentration is small, its ${\rm{pH}}$ is small and more than that of ${\rm{HCl}}$ .

Among the given aqueous solutions, ${\rm{Ba}}{\left( {{\rm{OH}}} \right)_2}$ , ${\rm{NaOH}}$ and ${{\rm{N}}_2}{{\rm{H}}_2}$ are the bases.

${\rm{Ba}}{\left( {{\rm{OH}}} \right)_2}$ is a strong base.

$\left[ {{\rm{Ba}}{{\left( {{\rm{OH}}} \right)}_2}} \right] = 0.1{\rm{ M}}$

On dissociation, it produces two moles of ${\rm{O}}{{\rm{H}}^ - }$ .

So,

$\begin{array}{l}\\\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 2 \times 0.1{\rm{ M}}\\\\{\rm{ = 0}}{\rm{.2 M}}\\\end{array}$

Since, the concentration is high, its ${\rm{pH}}$ is large.

${\rm{NaOH}}$ is a strong base.

$\left[ {{\rm{NaOH}}} \right] = 0.1{\rm{ M}}$

On dissociation, it produces one mole of ${\rm{O}}{{\rm{H}}^ - }$ .

So,

$\begin{array}{l}\\\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] = 1 \times 0.1{\rm{ M}}\\\\{\rm{ = 0}}{\rm{.1 M}}\\\end{array}$

Since, the concentration is high, its ${\rm{pH}}$ is large but less than that of ${\rm{Ba}}{\left( {{\rm{OH}}} \right)_2}$ .

${{\rm{N}}_2}{{\rm{H}}_2}$ is a weak base.

Its dissociation is not complete.

So,

$\left[ {{\rm{O}}{{\rm{H}}^ - }} \right] < {\rm{0}}{\rm{.1 M}}$

Since, the concentration is small, its ${\rm{pH}}$ is small and less than that of ${\rm{NaOH}}$ .

The order of the decreasing ${\rm{pH}}$ is as follows:

${\rm{Ba}}{\left( {{\rm{OH}}} \right)_2} > {\rm{NaOH}} > {{\rm{N}}_2}{{\rm{H}}_2} > {\rm{HOCl}} > {\rm{HCl}}$

Ans:

Therefore, the order of the decreasing ${\rm{pH}}$ is as follows:

${\rm{Ba}}{\left( {{\rm{OH}}} \right)_2} > {\rm{NaOH}} > {{\rm{N}}_2}{{\rm{H}}_2} > {\rm{HOCl}} > {\rm{HCl}}$