Question

A solution was made in a 200.00 mL volumetric flask consisting of 18.00 mL of 2.125 M HCl and 0.4104 grams of CaCO3, followed by adding enough water to bring the final volume up to the mark on the flask. A 15.00 mL sample of the solution was titrated with 0.1111 M NaOH to the equivalence point. How many mL of NaOH are used

Answer 1

16.2mL of 0.1111M NaOH are used

Step-by-step explanation:

volumetric flask of 250.0mL

As first, the HCl reacts with CaCO3 as follows:

2HCl + CaCO3 → H2O + CaCl2 + CO2

Where 2 moles of HCl react with 1 mole of CaCO3

To solve this question we must find the moles of each reactant in order to determine the moles of HCl that remains:

Moles HCl:

0.01800L * (2.125mol / L) = 0.03825 moles HCl

Moles CaCO3 -Molar mass: 100.09g/mol-

0.4104g * (1mol / 100.09g) = 0.00410 moles CaCO3

Moles HCl that reacts:

0.00410 moles CaCO3 * (2 mol HCl / 1 mol CaCO3) =

0.00820 moles HCl

Moles HCl that remains:

0.03825 moles HCl - 0.00820 moles HCl:

0.03005 moles HCl remains

The HCl reacts with NaOH as follows:

HCl + NaOH → H2O + NaCl

That means, to reach the equivalence point, the moles of NaOH that must be added = Moles HCl, in a sample of 15.00mL, the moles of HCl are:

0.03005 moles HCl * (15.0mL / 250.0mL) = 0.001803 moles HCl = Moles NaOH. The volume of 0.1111M NaOH is:

0.001803 moles NaOH * (1L / 0.1111moles) = 0.016L 0.1111M NaOH are required, that is:

16.2mL of 0.1111M NaOH are used