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?

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

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explanation:

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