Question

A) A volume of 100 mL of 1.00 M HCl solution is titrated with 1.00 M NaOH solution. You added the following quantities of 1.00 M NaOH to the reaction flask. Classify the following conditions after the equivalence point. Write down the items to their respective table.
i) 5.00 mL of 1.00 M NaOH

ii) 50.0 mL of 1.00 M NaOH

iii) 100 mL of 1.00 M NaOH

iv) 150 mL of 1.00 M NaOH

v) 200 mL of 1.00 M NaOH

Before equivalence point At equivalence point After equivalence point

Answer 1

Answer:

• i) 5.00 mL of 1.00 M NaOH: before the equivalence point

• ii) 50.0 mL of 1.00 M NaOH: before the equivalence point

• iii) 100 mL of 1.00 M NaOH: at the equivalence point

• iv) 150 mL of 1.00 M NaOH: after the equivalence point

• v) 200 mL of 1.00 M NaOH: after the equivalence point

Explanation:

1. First calculate the number of mol acid in the 100 mL of 1.00 M HCl solution.

Equation:

• Molarity = numbrer of moles of solute / volume of the solution in liters.

Thus,  you need to convert each volume from mL to liters, which is done dividing by 1,000.

Naming M the molarity, n the number of moles of solute (acid or base), and V the volume in liters:

$M=n/v\implies n=M\times V=1.00M\times 0.100L=0.100mol$

2. Now calculate the number of moles of NaOH for every condition (addition)

i) 5.00 mL of 1.00 M NaOH

$n=0.00500liter\times 1.00M=0.00500molNaOH$

Since the number of moles of NaOH added (0.00500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

ii) 50.0 mL of 1.00 M NaOH

$n=0.0500liter\times 1.00M=0.0500molNaOH$

Since the number of moles of NaOH added (0.0500mol) is less than the number of moles of acid in the solution (0.100mol), this is before equivalence point.

iii) 100 mL of 1.00 M NaOH

$n=0.100liter\times 1.00M=0.100molNaOH$

Since the number of moles of NaOH added (0.100mol) is equal to the number of moles of acid in the solution (0.100mol), this is at the equivalence point.

iv) 150 mL of 1.00 M NaOH

$n=0.150liter\times 1.00M=0.150molNaOH$

Since the number of moles of NaOH added (0.150mol) is greater than the number of moles of acid in the solution (0.100mol), this is after the equivalence point.

v) 200 mL of 1.00 M NaOH

This is more volume of NaOH, then this is also after the equivalence point.