Calculate Ka of acetic acid if its 0.05 M solution has molar conductivity of 7.814 × 10^-4Ω ^-1 m^2 mol^-1 at 25^∘C. Given : Λ m^∘ for CH3COOH = 3.907 × 10^-2Ω ^-1 m^2 mol^-1.
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![Calculating [H+] and pH from Ka](https://www.mi.mun.ca/users/pfisher/chemistry1011_134/img013.gif "Calculating [H+] and pH from Ka")
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![SOLVED: Calculate Ka and pKa of the acid using pH and molarity.moles unknown acid = 0.001215 molar mass of acid = 172.84molarity = 0.243 mol/LpH= 2.06kA= [A-][H3O+] / [HA]please include a rice](https://cdn.numerade.com/ask_previews/f6745ba3-6b77-4e7a-9de2-7d980958d194_large.jpg "SOLVED: Calculate Ka and pKa of the acid using pH and molarity.moles unknown acid = 0.001215 molar mass of acid = 172.84molarity = 0.243 mol/LpH= 2.06kA= [A-][H3O+] / [HA]please include a rice")
SOLVED: Calculate Ka and pKa of the acid using pH and molarity.moles unknown acid = 0.001215 molar mass of acid = 172.84molarity = 0.243 mol/LpH= 2.06kA= [A-][H3O+] / [HA]please include a rice
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![Calculate Ka of acetic acid from equilibrium concentration given below: [H(3)O^(+)]=[CH(3)COO^(-)]=1.34xx10^(-3)M, [CH(3)COOH]=9.866xx10^(-2)M](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/161343719_web.png "Calculate Ka of acetic acid from equilibrium concentration given below: [H(3)O^(+)]=[CH(3)COO^(-)]=1.34xx10^(-3)M, [CH(3)COOH]=9.866xx10^(-2)M")
Calculate Ka of acetic acid from equilibrium concentration given below: [H(3)O^(+)]=[CH(3)COO^(-)]=1.34xx10^(-3)M, [CH(3)COOH]=9.866xx10^(-2)M
