Solution of an acid and it's anion (that is it's conjugate base) or of a base and it's common cation are buffered. When we add a small amount of acid or base to any one of them, the pH of solution changes very little pH of buffer solution can be compoted as , for , acidic , buffer pH = pK_(a) + "log"(["Conjugate base"])/(["Acid"]) for basuc buffer pOH = pK_(b) + "log" (["Conjugated base"])/(["base"]) it is generally accepteed that a solution has useful buffer capacity (pH change resistng power) provided that the value of [salt or conjugate base]/[acid] for acidic acid buffer lies within the range of 1:10 to 10:1. Buffer capacity is max. When [conjugate base]=[acid] One litre of an aqueous solution contaion 0.15 mole of CH_(3)COOH(pK_(a) = 4.8) and 0.15 mole of CH_(3)COONa. After the addition of 0.05 moe of solid NaOH to this solution, the pH will be:

Solution of an acid and it's anion (that is it's conjugate base) or of

1.	Solution of an acid and it's anion (that is it's conjugate base) or of a base and it's common cation are buffered. When we add a small amount of acid or base to any one of them, the pH of solution changes very little pH of buffer solution can be compoted as , for , acidic , buffer pH = pK_(a) + "log"(["Conjugate base"])/(["Acid"]) for basuc buffer pOH = pK_(b) + "log" (["Conjugated base"])/(["base"]) it is generally accepteed that a solution has useful buffer capacity (pH change resistng power) provided that the value of [salt or conjugate base]/[acid] for acidic acid buffer lies within the range of 1:10 to 10:1. Buffer capacity is max. When [conjugate base]=[acid] One litre of an aqueous solution contaion 0.15 mole of CH_(3)COOH(pK_(a) = 4.8) and 0.15 mole of CH_(3)COONa. After the addition of 0.05 moe of solid NaOH to this solution, the pH will be:
Answer» `4.5` `4.8` `5.1` `5.4` Solution :`{:(CH_(3)COOH,,CH_(3)COONa),(I:""0.15,,""0.15):}` after addition of `0.05` MOLES of `NaOH` `{:(CH_(3)COOH,+,NaOH,RARR,CH_(3)COONa),(0.05,,0.05,,0.05):}` LEFT finally `{:(CH_(3)COOH,rarrCH_(3)COONa),(0.1 ""0.2,):}` `:. pH = 4.8 + "log" (0.2)/(0.1) = 5.1`