
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
164.1k+ views
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.
\[pOH = - log\left[ {O{H^-}} \right]\]
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Reservation Criteria 2025: SC, ST, EWS, and PwD Candidates

JEE Mains 2025 Cut-Off GFIT: Check All Rounds Cutoff Ranks

Other Pages
NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Thermodynamics Class 11 Notes: CBSE Chapter 5

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates

Neet Cut Off 2025 for MBBS in Tamilnadu: AIQ & State Quota Analysis

Karnataka NEET Cut off 2025 - Category Wise Cut Off Marks
