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What is the amount of potassium acetate to be added to 1 litre of N/10 acetic acid to make a buffer solution of pH=4 using acetic acid and potassium acetate? The dissociation constant of acetic acid =2*10^-5
let the amount of potassium acetate = m grams
M(CH3COOK) = m/92 .........1
normality of acetic acid = molarity*1
M( CH3COOH) = 0.1
Ka = [salt][H+]/[acid]
PH =4 means [H+] = 10-4
Ka = M(CH3COOK][H+]/[CH3COOH]
2*10-5 = [m/92][10-4]/[10-1]
m= 1.84grams
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