Answer
Verified
394.8k+ views
Hint: To get the concentration of hydroxide ions, pOH calculation can be done in the terms of dissociation constant and the molarity of the solution. Dissociation constant is the ratio of dissociated product ions to reactant ions.
Complete step-by-step solution:
Aniline hydrochloride is a salt that reacts with water and dissociates into aniline, which is a weak base. It is also a weak acid having aniline as its conjugate base. Concentration of hydroxide ion depends on the molarity of the solution. It is used in the titration process against strong bases like sodium hydroxide.
\[{C_6}{H_5}N{H_2} + {H_2}O \rightleftharpoons {C_6}{H_5}N{H_3}^ + + O{H^ - }\]
The equilibrium constant for this reaction is known as base-dissociation constant for the amine, denoted by \[{K_b}\]. pOH is determined to get the concentration of hydroxide ions as it is the negative algorithm of hydroxide ion concentration. It can also be written in the form of dissociation constant as-
\[pOH = \dfrac{1}{2}[p{K_b} - \log b]\]
We are given the value of \[{K_b}\], so using the below formula and putting its value in it, we get the value of \[p{K_b}\].
\[p{K_b} = - \log {K_b}\]
\[ = - \log (4.0 \times {10^{ - 10}})\]
\[ = 9.34\]
Now we can put this value in the below formula and get the value of \[pOH\]
\[pOH = \dfrac{1}{2}[p{K_b} - \log b]\]
\[pOH = \dfrac{1}{2}[9.34 - \log (0.01)]\]
\[\therefore pOH = 4.67 - ( - 2)\]
\[\therefore pOH = 6.67\]
\[\therefore - \log [O{H^ - }] = 6.67\]
\[\therefore [O{H^ - }] = 4.7 \times {10^6}\]
From this we conclude that it is a weak base as its pH comes out to be 7.33 i.e. on subtracting pOH from 14.
Note: There is another method to solve this numerical. This method involves ionisation constant and change in concentration from initial to final for all the species using formulas like \[{K_\alpha } = \dfrac{{{K_w}}}{{{K_b}}}\] and \[{K_\alpha } = c{\alpha ^2}\], where \[\alpha < < 1\].
Complete step-by-step solution:
Aniline hydrochloride is a salt that reacts with water and dissociates into aniline, which is a weak base. It is also a weak acid having aniline as its conjugate base. Concentration of hydroxide ion depends on the molarity of the solution. It is used in the titration process against strong bases like sodium hydroxide.
\[{C_6}{H_5}N{H_2} + {H_2}O \rightleftharpoons {C_6}{H_5}N{H_3}^ + + O{H^ - }\]
The equilibrium constant for this reaction is known as base-dissociation constant for the amine, denoted by \[{K_b}\]. pOH is determined to get the concentration of hydroxide ions as it is the negative algorithm of hydroxide ion concentration. It can also be written in the form of dissociation constant as-
\[pOH = \dfrac{1}{2}[p{K_b} - \log b]\]
We are given the value of \[{K_b}\], so using the below formula and putting its value in it, we get the value of \[p{K_b}\].
\[p{K_b} = - \log {K_b}\]
\[ = - \log (4.0 \times {10^{ - 10}})\]
\[ = 9.34\]
Now we can put this value in the below formula and get the value of \[pOH\]
\[pOH = \dfrac{1}{2}[p{K_b} - \log b]\]
\[pOH = \dfrac{1}{2}[9.34 - \log (0.01)]\]
\[\therefore pOH = 4.67 - ( - 2)\]
\[\therefore pOH = 6.67\]
\[\therefore - \log [O{H^ - }] = 6.67\]
\[\therefore [O{H^ - }] = 4.7 \times {10^6}\]
From this we conclude that it is a weak base as its pH comes out to be 7.33 i.e. on subtracting pOH from 14.
Note: There is another method to solve this numerical. This method involves ionisation constant and change in concentration from initial to final for all the species using formulas like \[{K_\alpha } = \dfrac{{{K_w}}}{{{K_b}}}\] and \[{K_\alpha } = c{\alpha ^2}\], where \[\alpha < < 1\].
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE