will 0.04 M CH3COOH and 0.2M NaOH form buffer solution?? as equivalents of the strong base are more in this case and buffer consists of weak acid and salt of weak acid and strong base??
as u hav not not mention volume i'm taking 1 ltr if u use buffer equation here ph= pka + log [salt] / [acid -salt ] it would not yield any result becoz conc of acid is less than base quantity under log is negative n log is not define for negative values so it is not buffer ...........
Moderation Team
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