How much `CH_3COONa` should be added to 1 litre of 0.01 M ` CH_3 COOH` to make . A buffer of pH = 4.1 ?
` (K_a " for " CH_3COOH = 1.8 xx 10 ^(-5))`

Explanation

Given [Acid ] `= 0.01 M, [" Salt" ] =? `
` K_a -= 1.8 xx 10 ^(-5) , pH =4.1 `
According to the Henderson.s equation
` " " pH =pK_a +log_(10) (["Salt"])/([" Acid"]) `
Substituting the values , we have
` " " 4.1 = - log _(10 ) 1. 8 xx 10 ^(-5) +log_(10 ) ([" Salt"])/( 0.01) `
or , ` " " 4.1 = 4. 745 + log _(10 ) ([" Salt "])/( 0.01) `
or, ` " " log_(10 ) (["Salt"])/(0.01 ) = 4.1 - 4.75 = - 0.645`
or , ` " " (["Salt "])/( 0.01 ) = " antilog "_(10) (-0.645) = 0.226`
` therefore " " [" Salt" ] = 0.226 xx 0.01 `
` " " = 2.26 xx 10 ^(-3) " mol " L^(-1) `
The mass of `CH_3 COONa` to be added to 1 litre
` " " = 2.26 xx 10 ^(-3) xx 82 `
` " " ( because ` molecular mass of ` CH_3 COONa = 82 ) `
= 0 . 185 g