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How many milliliters of 0. 120 m naoh are required to titrate 50. 0 ml of 0. 0998 m butanoic acid to the equivalence point? the ka of butanoic acid is 1. 5 × 10-5.

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The volume of NaOH required to titrate 50 mL of 0.0998 M butanoic acid to the equivalence point is 41.58 mL

Balanced equation

C₄H₈O₂ + NaOH —> C₄H₇O₂Na + H₂O

From the balanced equation above,

  • The mole ratio of the acid, C₄H₈O₂ (nA) = 1
  • The mole ratio of the base, NaOH (nB) = 1

From the question given above, the following data were obtained:

  • Concentration of base, NaOH (Cb) = 0.12 M
  • Concentration of acid, C₄H₈O₂ (Ca) = 0.0998 M
  • Volume of acid, C₄H₈O₂ (Va) = 50 mL
  • Volume of base, NaOH (Vb) =?

CaVa / CbVb = nA / nB

(0.0998 × 50) / (0.12 × Vb) = 1

4.99 / (0.12 × Vb)  = 1

Cross multiply

0.12 × Vb = 4.99

Divide both side by 0.12

Vb = 4.99 / 0.12

Vb = 41.58 mL

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