Question

3.0 M HI Unknown NaOH initial burette reading 0.2 ml 0.7 ml final burette reading 47.6 ml 37.5 ml What is the concentration of the base (NaOH) in the following titration?

Answer 1

To calculate the concentration of NaOH in a titration with HI, subtract the initial burette reading from the final one to find the volume of NaOH used, then use the moles of HI (from its concentration and volume) and the volume of NaOH to find the concentration of NaOH.

Step-by-step explanation:

To determine the concentration of the base (NaOH) in the titration with 3.0 M HI, we can use the information provided. First, we need to find the volume of NaOH solution used in the titration. This is done by subtracting the initial burette reading from the final burette reading. For instance, if the initial reading was 0.7 mL and the final reading was 37.5 mL, the volume used would be 37.5 mL - 0.7 mL = 36.8 mL. Knowing the concentration of HI and the volume of HI used, we can calculate the moles of HI, which will be equal to the moles of NaOH at the equivalence point due to the 1:1 mole ratio in the reaction HI + NaOH → NaI + H2O.

For example, if 25.00 mL of HI was used, the moles of HI would be concentration × volume = 3.0 M × 0.025 L = 0.075 moles. Since the moles of NaOH are the same, the concentration of NaOH can then be calculated using the formula concentration = moles / volume where volume is converted to liters. So if 36.8 mL of NaOH was used, that would be 0.075 moles / 0.0368 L = approximately 2.038 M, which is the concentration of NaOH.

Answer 2

This is a neutralization, which means we are mixing a base with an acid until the mixture becomes neutral. We add more HI to the NaOH until the number of acid equivalents is equal to the number of base equivalents. We can calculate the acid equivalents using normality and volume, and the same with base equivalents.

A = acid equivalents = normality*volume (in the acid solution)

B = base equivalents = normality*volumen (in the base solution)

A = B

NA*VA = NB*VB

HI is an acid which releases only 1 acid equivalent per molecule, so its molarity is equal to its normality.

NaOH is a base which releases only 1 base equivalent per formula unit, so its molarity is equal to its normality.

MA*VA = MB*VB

We’re trying to find out NaOH molarity, which is equal to the NaOH normality.

MB = MA*VA/VB

DATA:

MA = 3.0 M

The volumen of HI used can be calculated by subtracting the final volume of the burette to its initial volume (the final volume is smaller, as we have taken some volume away)

VA = V0-Vf = 0.7 ml - 0.2 ml = 0.5 ml

VB = V0-Vf = 47.6 ml - 37.5 ml = 10.1 ml

MB = 3.0M*0.5M/10.1M = 0.149 M