Question

Arrange the following solutions in the decreasing order of their pOH:
(A) 0.01 M HCl
(B) 0.01 M NaOH
(C) 0.01 M \[C{H_3}COONa\]
(D) 0.01 M NaCl
A. \[\left( A \right) > \left( C \right) > \left( D \right) > \left( B \right)\]
B. \[\left( B \right) > \left( D \right) > \left( C \right) > \left( A \right)\]
C. \[\left( B \right) > \left( C \right) > \left( D \right) > \left( A \right)\]
D. \[\left( A \right) > \left( D \right) > \left( C \right) > \left( B \right)\]

Answer 1

Hint: pOH and pH illustrate the acidic or basic an aqueous solution. pOH of an aqueous solution is the negative logarithm of the concentration of hydroxide ion (\[O{H^-}\]) concentration.
\[pOH = - log\left[ {O{H^-}} \right]\]
It is the reverse of pH which illustrates the acidity of a substance.

Formula Used:
\[pOH{\rm{ }} = {\rm{ }} - log\left[ {O{H^-}} \right]\]
\[pH = - log\left[ {{H^ + }} \right]\]
\[pH + pOH = 14\]; where \[\left[ {O{H^-}} \right],{\rm{ }}\left[ {{H^ + }} \right]\] are the concentrations of hydroxyl ions and hydrogen ions respectively.

Complete Step by Step Answer:
For the given question, we have to find out the pOH of the given solutions.
A. 0.01 M HCl
HCl dissociates to form one mole of \[{H^ + }\] and one mole of \[C{l^ - }\] ion.
So, the \[\left[ {{H^ + }} \right] = 0.01M\]
Thus, the \[pH = - log\left( {{{10}^{ - 2}}} \right) = 2\]
Putting the value of pH in the below equation we get,
As we know that\[pH + pOH = 14\]
pOH = 14-2
pOH = 12

B. 0.01 M NaOH
NaOH dissociates to form one mole of\[N{a^ + }\] and one mole of \[O{H^ - }\] ion.
So, the \[\left[ {O{H^-}} \right] = 0.01M\]
So, \[pOH = - log\left( {{{10}^{ - 2}}} \right) = 2\]

C. 0.01 M \[C{H_3}COONa\]
Its chemical name is Sodium acetate.
It is the salt of acetic acid and sodium hydroxide i.e., it is the salt of a weak acid and a strong base.
So, it is alkaline in nature.
So, its pH will be less than 7 and as a result its pOH is higher than 7.

D. 0.01 M NaCl
It is a salt of strong acid, HCl and strong base, NaOH.
So, it is neutral i.e., neither acidic nor basic.
Its pH and pOH is 7.
So, the order of decreasing pOH is \[\left( A \right) > \left( D \right) > \left( C \right) > \left( B \right)\].
So, option D is correct.

Additional Information: We know that water molecules have weak ionisation forming conjugate acid-base pairs i.e., hydrogen ion and hydroxyl ion in the solution.
\[{H_2}O \mathbin{\lower.3ex\hbox{$\buildrel\textstyle\rightarrow\over{\smash{\leftarrow}\vphantom{_{\vbox to.5ex{\vss}}}}$}} {H^ + } + O{H^ - }\]
The equilibrium constant for this reaction is:-
\[K = \frac{{\left[ {O{H^-}} \right]\left[ {{H^ + }} \right]}}{{\left[ {{H_2}O} \right]}}\]
where
\[\left[ {O{H^-}} \right],{\rm{ }}\left[ {{H^ + }} \right]\]and \[\left[ {{H_2}O} \right]\] are the concentrations of hydroxyl ions, hydrogen ions, and water respectively.
The concentration of water can be assumed to be a constant.
Therefore,
where \[{K_w}\] is the ionic product of water.
So, \[{K_w} = \left[ {O{H^-}} \right]\left[ {{H^ + }} \right]\]
Taking log both sides we get,
\[log{K_w} = log\left[ {O{H^-}} \right] + log\left[ {{H^ + }} \right]\]
\[ \Rightarrow log{K_w} = pOH + pH\]
The ionic product of water is a constant and has the value of \[{10^{ - 14}}\].
Putting this value in the above equation we get, pH + pOH=14.

Note: An insufficient pOH value implies increased basicity or alkalinity, while an elevated pOH value demonstrates increased acidity. This is because a low value of pOH means a low concentration of hydroxyl ions which means a high concentration of hydrogen ions. So, the solution is acidic. Similarly, the high value of pOH indicates a higher concentration of hydroxyl ions thus high alkalinity of the solution. Likewise, a low pH demonstrates elevated acidity, while an increased pH implies high basicity.