20 mL of 0.1 M acetic acid is mixed with 50 mL of potassium acetate. K_ of acetic acid =1.8 10^ -5 at 27^ C . Calculate the concentration of potassium acetate if pH of mixture is 4.8 .

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Correct answer: B. 0.04 M

For an acidic buffer, pH = p K_ + [ Salt ] [ Acid ] (i) Let the concentration of potassium acetate solution =x M 20 mL of 0.1 M acetic acid =20 0.1 millimol =2 millimol 50 mL of x M potassium acetate =x 50 millimol aligned & =50 x milimol \ pH & =4.8 (Given) aligned Substitute all value in eq. (i), aligned 4.8 & =- ( 1.8+10^ -5 )+ 50 x 2 \ 25 x & =0.06 \ x & =0.04 M aligned

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