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```
The pH of a solution obtained by mixing 100 ml of 0.2 M CH 3 COOH with 100 ml of 0.2 M NaOH will be (pKa fro CH 3 COOH = 4.74 and log 2 = 0.301) (Ans : 8.87)

```
3 years ago

Arun
25768 Points
```							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
```
3 years ago
Sunaina
16 Points
```							Its wrong becoz the formula is wrong,, ,it will be 7+(pka-logc)                ..........lekin tumne wrong  formula lga k ans nikal diya h
```
one year ago
desu
13 Points
```							to tumne konsa sahi formula lga liya 😂😂😂...its 7+1/2 (pka+logC) dude... hope you get it right now...
```
one year ago
Yash Chourasiya
256 Points
```							Dear StudentWe 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)=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]=8.87I Hope this answer will help you.Thanks & RegardsYash Chourasiya
```
6 months ago
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