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