tonydipiop8r6n0 tonydipiop8r6n0
  • 15-05-2018
  • Chemistry
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If the ph of hc3h5o2 is 4.2 and the ka 1.34x10^-5, what is the equilibrium concentration

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ChemistryHelper2024
ChemistryHelper2024 ChemistryHelper2024
  • 24-05-2018
CH₃CH₂COOH + H₂O ↔  CH₃CH₂COO⁻ + H₃O⁺ 

pH = 0.5 pKa + 0.5 pCa
0.5 pCa = pH - 0.5 pKa
              = 4.2 - (0.5 * (-log 1.34 x 10⁻⁵)) = 1.76
pCa = 3.53
Ca = antilog - 3.52 = 3 x 10⁻⁴ 
where Ca is the acid concentration 
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