3.0 g of pure acetic acid and 4.1 g of anhydrous sodium acetate are dissolved together in water and the solution is made up to 500 ml. Calculate the pH of the solution. Given `K_(a)` of acetic acid is `1.75xx10^(-5)`.

Explanation

Calculation of the concentration of acetic acid : mass/`dm^(3)=Nxxg` eq. mass
`"mass"//500cm^(3)=(Nxxg" eq. mass")/2rArrN_(CH_(3)COOH)=("Mass"//500cm^(3)xx2)/("g.eq.mass")`
Molecular wt. of `CH_(3)COOH=12xx2+1xx4+16xx2=24+04+32=60`
`[CH_(3)COOH]=(3.0xx2)/60=6.0/60=0.1N`
Calculation of concentration of sodium acetate. `"mass"//500cm^(3)=(Nxxg" eq.mass")/2rArrN_(CH_(3)COONa)=("Mass"//500cm^(3)xx2)/("g.eq.mass")`
Molecular wt. of `[CH_(3)COONa]=12xx2+1xx3+16xx2+1xx23=82`
`[CH_(3)COONa]=(41xx2)/82=8.2/82=0.1N`
`pH=pK_(a)+"log"(["Salt"])/(["Acid"])rArrpH=-log_(10)K_(a)+"log"([CH_(3)COONa])/([CH_(3)COOH])`
`pH=-log_(10)(1.75xx10^(-5))+"log"([0.1])/([0.1])=-log1.75-log10^(-5)+log1`
`pH=-0.2430+5+0=4.7570`