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Question: Acetic acid \[\dfrac{N}{10}\] was titrated with \[\dfrac{N}{10}\] NaOH. When 25%, 50% and 75% of tit...

Acetic acid N10\dfrac{N}{10} was titrated with N10\dfrac{N}{10} NaOH. When 25%, 50% and 75% of titration is over then the pH of the solution will be : [Ka=10−5{{K}_{a}}={{10}^{-5}}]
(A) 5+log⁡13,5,5+log⁡35+\log \dfrac{1}{3},5,5+\log 3
(B) 5+log⁡3,4,5+log⁡135+\log 3,4,5+\log \dfrac{1}{3}
(C) 5−log⁡13,5,5−log⁡35-\log \dfrac{1}{3},5,5-\log 3
(D) 5−log⁡13,4,5+log135-\log \dfrac{1}{3},4,5+\text{log}\dfrac{1}{3}

Explanation

Solution

Acetic acid is an organic compound which is a carboxylic acid attached with a methyl group. It is produced industrially by the carbonylation of methanol. It is a colourless liquid with a pungent smell. The conjugate base of acetic acid is acetate. The acid dissociation constant of acetic acid is 4.76.

Complete step by step solution:
The nf{{n}_{f}} of both acetic acid and sodium hydroxide is equal to 1. So accordingly the normality and molarity will be equal to each other. So,
N10=110=0.1MAceticacid\dfrac{N}{10}=\dfrac{1}{10}=0.1M Acetic acid and 0.1M NaOH.
The reaction for acetic acid and sodium hydroxide is the following:
CH3COOH+NaOH→CH3COO−Na++H2OC{{H}_{3}}COOH+NaOH\to C{{H}_{3}}CO{{O}^{-}}N{{a}^{+}}+{{H}_{2}}O

So at t=0 the molar concentration of acetic acid and sodium hydroxide is 0.1
At t2{{t}_{2}} =25% the molar concentration of acid will be = 0.1−0.025=0.0750.1-0.025=0.075 M and for salt will be = 0.1−0.025=0.0750.1-0.025=0.075 M
At t3{{t}_{3}} =50% the molar concentration of acid will be = 0.1−0.050=0.0500.1-0.050=0.050 M and for salt will be= 0.1−0.050=0.0500.1-0.050=0.050 M
At t4{{t}_{4}} =75% the molar concentration of acid will be = 0.1−0.075=0.0750.1-0.075=0.075 and for salt will be= 0.1−0.075=0.0750.1-0.075=0.075 M
Now the formula for calculating pH is = −log⁡ka+log⁡[salt][acid]-\log {{k}_{a}}+\log \dfrac{[salt]}{[acid]}

Now we will firstly calculate −log⁡ka-\log {{k}_{a}}. So,
−log⁡ka=−log⁡10−5-\log {{k}_{a}} = -\log {{10}^-5}
⇒−log⁡ka=+5log⁡10=5\Rightarrow -\log {{k}_{a}}=+5\log 10=5
So now we will calculate the pH for 25% completion: pH=−log⁡ka+log⁡[0.025][0.075]pH=-\log {{k}_{a}}+\log \dfrac{[0.025]}{[0.075]}
⇒[pH=5+log⁡13]\Rightarrow [pH=5+\log \dfrac{1}{3}]
So now we will calculate the pH for 50% completion pH=−log⁡ka+log⁡[0.050][0.050]pH=-\log {{k}_{a}}+\log \dfrac{[0.050]}{[0.050]}
⇒[pH=5+log⁡1=5]\Rightarrow [pH=5+\log 1=5]
So now we will calculate the pH for 75% completion pH=−log⁡ka+log⁡[0.075][0.025]pH=-\log {{k}_{a}}+\log \dfrac{[0.075]}{[0.025]}
⇒[pH=5+log⁡3]\Rightarrow [pH=5+\log 3]

So the correct answer is option (A).

Note: The acetic acid has some antibacterial qualities due to which it is used as an antiseptic. It helps in manufacturing of rayon fibre. It used to treat cancer. It helps in manufacturing of rubber. In liquid state it is used as a hydrophilic and acts as a protic and polar solvent. It is regarded as an industrial solvent.