The volume of the solution that the student will need to add to reach the equivalence point is 11.6 ml of 0.1400M NaOH solution.
For this reaction, 0.154 g of chloroacetic acid was diluted to 250 ml and titrated with 0.1400 M NaOH solution.
The reaction between chloroacetic acid ClCH₂COOH and sodium hydroxide NaOH is given as:
ClCH₂COOH +NaOH → ClCH₂COO⁻ + Na⁺ + H₂O
Here, 1 mole of the acid will react with 1 mole of the base
so, the student will reach the equivalence point when
no.of moles of chloroacetic acid = no.of moles of sodium hydroxide
molar mass of ClCH₂COOH = 94.5 g
∴ 0.154 g × (1 mol/94.5 g) = 1.63 × 10⁻³ moles of ClCH₂COOH
so, to reach the equivalence point the student should add 1.63 × 10⁻³ moles of NaOH
∴ 1.63 × 10⁻³ moles of NaOH × (1L / 0.1400 moles NaOH )
=0.0116L of 0.1400 M NaOH
Thus, 11.6 ml of 0.1400M NaOH solution is required to reach the equivalence point.
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