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Determine the volume in mL of 0.242 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 46.79 mL of 0.204 M propanoic acid(aq). The Ka of propanoic acid is 1.3 x 10-5.

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The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL

Balanced equation

CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O

From the balanced equation above,

  • The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
  • The mole ratio of the base, NaOH (nB) = 1

How to determine the volume of NaOH

  • Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
  • Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
  • Molarity of base, NaOH (Mb) = 0.242 M
  • Volume of base, KOH (Vb) =?

MaVa / MbVb = nA / nB

(0.204 × 46.79) / (0.242 × Vb) = 1

Cross multiply

0.242 × Vb = 0.204 × 46.79

Divide both side by 0.242

Vb = (0.204 × 46.79) / 0.242

Vb = 39.44 mL

Thus, the volume of NaOH needed for the reaction is 39.44 mL

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