The volume of 0.0600 M NaOH solution the student will need to add to reach the equivalence point intitrationof 0.231 g of chloroacetic acid is 40.7 mL. This is determined by calculating the moles of chloroacetic acid present, and then calculating thevolumeof NaOH solution required to equal these moles, because the reaction is in a 1:1 ratio.To answer this question, we first need to know that theequivalence pointin a titration is reached when the amount of titrant added equals the amount of analyte present in the original sample. Chloroacetic acid (HCH2ClCO2) is a monoprotic acid and that means it will react with NaOH in a 1:1 ratio. Thus, we can calculate the moles ofchloroacetic acidfirst.The molar mass of chloroacetic acid is about 94.5 g/mol,  so there is 0.231 g/94.5 g/mol = 0.00244 mol of chloroacetic acid. Now, matching the moles of NaOH needed (since it's a 1:1 ratio), we can calculate the volume of 0.0600 M NaOH solution needed.To find the volume, we divide themolesof NaOH needed by its concentration in the solution: 0.00244 mol / 0.0600 M = 0.0407 L, or 40.7 mL.In conclusion, the student will need to add 40.7 mL of the 0.0600 M NaOH solution to reach the equivalence point in his titration of chloroacetic acid.To know more abouttitration,visit:brainly.com/question/31271061#SPJ11...

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