Quantitative Analysis This question concerns the titration of 10.0 mL of 0.0400

Question

Quantitative Analysis

This question concerns the titration of 10.0 mL of 0.0400 M NaOH with 0.0200 M HCl. Neglect activity coefficients and assume 25 °C. [Note: as shown on p. 22 of the Equation Guide, we define the “fraction of titration” as the added volume of titrant divided by the equivalence volume. Thus, = 0 at the beginning of the titration and = 1 at the equivalence point.]

b. (1 pt) What volume of titrant must be added to reach the equivalence point? Calculate V.eqv using an explicit mole ratio and units and labels throughout.

Explanation / Answer

Neutralisation reaction:

NaOH (aq.) + HCl (aq.) --------------------> NaCl (aq.) + H2O (l)

Number of moles of NaOH given = 0.0400 * 10.0 / 1000. = 0.000400 mol

From the above balanced equation,

1 mol of NaOH neutralises 1 mol of HCl

So, number of moles of HCl required = 0.000400 mol

Knowing the molarity formula,

Molarity = number of moles / volume of solution in Litres

0.0200 = 0.000400 / V

V = 0.000400 / 0.0200

V = 0.0200 L

Or

V = 20.0 mL

SO, volume of HCl solution required = 20.0 mL

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