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Grade: 12th pass
        
 
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)
                            I have gone through the answers to this question on askiitians.....but i dont get this part where the concentration of the salt is considered to be the half of the initial concentration of CH3COOH  and NaOH.I understand that it wont be equal to 0.2,but why exactly half ?
2 years ago

Answers : (1)

Arun
23346 Points
							
PH = 7 + 1/2 pKa + 1/2 log c
0.2 M in 100 ml = 20 equivalent
20 equivalent CH3COOH + 20 equivalent NaOH → 20 equivalent  CH3COONa
20 equivalent= 20/100 = 0.2
pH = 7+(4.74/2)+ 1/2 log 0.2
      =  7+ 2.37 + 1/2 log( 2x 10^-1)
     = 9.37 +  1/2 log2 - 1/2 log 10
     = 9.37 + (0.301/2) - 1/2
    = 9.02
hope this answer helps u
2 years ago
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