Chemistry: Back Titration

Question

50.9 mL of 1.16 M nitric acid is added to 29.9 mL of sodium hydroxide, and the resulting solution is found to be acidic.

17.6 mL of 1.10 M calcium hydroxide is required to reach neutrality.

What is the molarity of the original sodium hydroxide solution?

 

Answer 1

Step 1: Acidic Solution of HNO3 + NaOH

To find moles of Nitric Acid:

(0.0509 L)(1.16 mol/L) = 0.059 mol HNO3

To find moles of NaOH:

(0.0299 L)(x) = 0.0299x mol NaOH ––––> x = molarity of NaOH solution

We cannot yet determine the exact moles of NaOH , but we do know that adding basic NaOH to acidic HNO₃ still results in an acidic solution, so we know that <0.059 moles of NaOH were added. In step 2, the remaining moles of ^–OH will be added to completely neutralize the 0.059 mol of acid.

Step 2: Neutralization of Solution by Ca(OH)₂

To find moles of Ca(OH)₂:

(0.0176 L)(1.1 mol/L) = 0.019 mol Ca(OH)₂

To find moles of ^–OH in 0.019 mol Ca(OH)₂

0.019 mol Ca(OH)₂ x (2 mol ^–OH / 1 mol Ca(OH)₂) = 0.038 mol ^–OH

This step reaches neutralization. Keep in mind that by adding 1 mol of Ca(OH)₂, we actually add 2 molar equivalents of ^–OH. After adding 0.038 mol of ^–OH, we reach neutralization.

Neutralization: (moles of HNO₃) – (moles of ^–OH) = 0

(0.059 mol) – (0.0299x + 0.038) = 0

0.021 = 0.0299x

x = 0.70 M