Titration: Back Titration

Question

36.8 mL of 1.92 M perchloric acid is added to 43.3 mL of sodium hydroxide, and the resulting solution is found to be acidic.

26.7 mL of 0.693 M barium hydroxide is required to reach neutrality.

What is the molarity of the original sodium hydroxide solution?

Answer 1

We know that at neutrality the number of moles of acid and the number of moles of base have been added in equal amounts. Therefore, we need to find out how many moles of acid we have to begin with and how many moles of base we need for neutralization.

Since we are starting with 36.8 mL of HClO4 with a molarity of 1.92M,

36.8 mL * (1L / 1000mL) * (1.92 moles / 1L) = 0.070656 moles of HClO4

Now, we need to make sure we have the balanced equation.

HClO4 + NaOH --> H2O + Na+ + ClO4-

Since this is in a 1:1 molar ratio of acid to base we know that we need 0.070656 moles of base to neutralize our acid.

To find out how many moles of base were added after the initial part of the reaction (when it was still acidic)...

26.7mL * ( 1L / 1000mL) * (0.693 moles / L) = 0.0185031 moles of Ba(OH)2

Now, for every Ba(OH)2 there are two equivalents of base, so this is like adding 0.037 moles of OH-.

If we subtract the two amounts, the 0.0707 moles of base that we needed to add, and the 0.037 moles of base that we had to add at the end, we get 0.0337 moles of base that were added.

Now, finally, since we know that we had 43.4 mL ( 0.0434 L) and we know how many moles of base there were, we can find the molarity.

0.0337 moles of base / 0.0434 L = 0.776M