Question

A chemistry student weighs out 0.154 g of chloroacetic acid (HCH2CICO2) into a 250. mL volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1400 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point.

Answer 1

Answer:

11.6 mL of 0.1400 M of NaOH is required to reach equivalence point.

Explanation:

Chloroacetic acid is an monoprotic acid.

Neutralization reaction: $ClCH_{2}COOH+NaOH\rightleftharpoons ClCH_{2}COONa+H_{2}O$

So, 1 mol of chloroacetic acid is neutralized by 1 mol of NaOH.

Molar mass of chloroacetic acid = 94.5 g/mol

So, 0.154 g of chloroacetic acid = $\frac{0.154}{94.5}$ moles of chloroacetic acid

                                                     = 0.00163 moles of chloroacetic acid

Lets assume V mL of 0.1400 M of NaOH is required to reach equivalence point.

So, number of moles of NaOH needed to reach equivalence point

      = $\frac{0.1400\times V}{1000}$ moles

So, $\frac{0.1400\times V}{1000}=0.00163$

or, V = 11.6

Hence, 11.6 mL of 0.1400 M of NaOH is required to reach equivalence point.