Ncert Solutions Chemistry Class 11th

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Here, total meq of acetic acid = 50 × 0.1 = 5

And total meq of NaOH = 25 × 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH3COONa = 2.5

pH=pKa+log10 [S] [A]

pH=4.76+log102.52.5=4.76=476×10−2

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New answer posted•a year ago
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a year ago

Organic pollutants in water

 

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