This can be done using Henderson Hasselbalch equation
The equation is
color(blue)(pH=pK_(CH_3COOH)+log (([CH_3COONa])/([CH_3COOH])))pH=pKCH3COOH+log([CH3COONa][CH3COOH])
Let us consider that xmLxmL 0.1M sodium Acetate is to be mixed with (200-x)mL(200−x)mL 2M acetic acid solution to prepare 200ml or 0.2L buffer solution of pH=4.50pH=4.50
So [CH_3COONa]xx0.2=x xx10^-3xx0.1[CH3COONa]×0.2=x×10−3×0.1
And [CH_3COOH]xx0.2=(200-x) xx10^-3xx2[CH3COOH]×0.2=(200−x)×10−3×2
Hence ([CH_3COONa])/([CH_3COOH])=x/((200-x)xx20)[CH3COONa][CH3COOH]=x(200−x)×20
Again pK_(CH_3COOH)=4.75pKCH3COOH=4.75
Inserting these in Henderson Hasselbalch equation
color(blue)(pH=pK_(CH_3COOH)+log (([CH_3COONa])/([CH_3COOH])))pH=pKCH3COOH+log([CH3COONa][CH3COOH])
color(blue)(=>4.50=4.75+log (([CH_3COONa])/([CH_3COOH])))⇒4.50=4.75+log([CH3COONa][CH3COOH])
color(blue)(=>log (([CH_3COONa])/([CH_3COOH]))=4.5-4.75=-0.25)⇒log([CH3COONa][CH3COOH])=4.5−4.75=−0.25
color(blue)(=>log (x/((200-x)xx20))=-0.25)⇒log(x(200−x)×20)=−0.25
color(blue)(=>(x/((200-x)xx20))=10^(-0.25))⇒(x(200−x)×20)=10−0.25
color(blue)(=>(((200-x)xx20)/x)=10^(0.25))⇒((200−x)×20x)=100.25
color(blue)(=>(200/x-1)=1/20 xx10^(0.25))⇒(200x−1)=120×100.25
color(blue)(=>200/x=1+1/20 xx10^(0.25))⇒200x=1+120×100.25
color(red)(=>x=183.7ml)⇒x=183.7ml
So volume of CH_3COONa=183.7mlCH3COONa=183.7ml
And volume of CH_3COOH=(200-183.7)mL=16.3mLCH3COOH=(200−183.7)mL=16.3mL
