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# The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH will be (pKa fro CH3COOH = 4.74 and log 2 = 0.301)(Ans : 8.87)

Arun
25763 Points
3 years ago
pH=pKa....+ …......log(salt/acid)As we are mixing equal volumes of the two solution, so the final concentration of NaOH as well as of acid would be ½ of the initial concentraton.Hence, pH=4.74+....log(0.1/0.1)=pKaHere we have assumed that the strong base dissociates completely(in general it is true), but the dissociation of weak acid is quite low.

pH = 1/2[pKw + pKa + log C]
= 1/2[14 +4.74 + log 0.1]
8.87
Sunaina
16 Points
2 years ago
Its wrong becoz the formula is wrong,, ,it will be 7+(pka-logc)                ..........lekin tumne wrong  formula lga k ans nikal diya h
desu
13 Points
2 years ago
to tumne konsa sahi formula lga liya 😂😂😂...its 7+1/2 (pka+logC) dude... hope you get it right now...
Yash Chourasiya
11 months ago
Dear Student

We can calculate this by using the formula :7+1/2 (PKa+logC)
As we are mixing equal volumes of the two solution, so the final concentration of NaOH as well as of acid would be ½ of the initial concentraton.
Hence, pH=4.74+....log(0.1/0.1)=pKa
Here we have assumed that the strong base dissociates completely(in general it is true), but the dissociation of weak acid is quite low.

pH= 1/2[pKw +pKa + log C]
= 1/2[14 +4.74 + log 0.1]
=8.87