When performing acid-base titrations, the first step is often a standardization

Question

When performing acid-base titrations, the first step is often a standardization of the titrant, which is an accurate Determination of its concentration. Acidic titrants might be standardized in titration of a well-defined amount of a solid base, such as sodium carbonate. In such a procedure, why is it not necessary to know the exact amount of water used in preparation of a solution of Na2CCO3 used in standardization of an acid? The standardization of an acid might also be carried out using a solution of a base with known concentration. For example, it is found that 37.60 ml of 0.210 M NaOH are required to neutralize 25.05 ml of H2SO4 solution in a titration experiment. Calculate the molarity and normality of the H2SO4 solution. In the example given above, the amount of acid to be titrated has to be delivered by a pipet so that its volume is accurately known. However, the solution of the acid is often diluted to facilitate the titration process. Explain why additional de-ionized water might be added without affecting the results of the titration.

Explanation / Answer

point 1: some reagents when exposed to air in solutions got oxidised or reduced due to the prsence of actve species in the air. so the solutions are always titrted with another regent to find its original strength at that point. so in ths contest it is not necessary to know the exact conc. of Na2CO3

point 2:

2NaOH + H2SO4 > Na2SO4 + 2H2O

This tells us we have a 2:1 mole to mole ratio. Now since M=mol/L, we can calculate the moles of NaOH.

0.210M x .0376L = 0.007896 moles NaOH.

We can use the mole to mole ratio from the balanced equation to move to moles of H2SO4.

0.007896 mol NaOH x (1 mol H2SO4/2 mol NaOH) = 0.003948 mol H2SO4.

Using these moles of H2SO4, we can then go back to the equation M=mol/L, and solve for M for H2SO4.

0.003948 mol H2SO4/0.02505L H2SO4 = 0.158 M H2SO4.

Normality = molarity/basicity

Normality = 0.158M / 2 = 0.079 N

3) If more concentrated solution is titrated against a base lead to the error at the end point even if thesmall excess base is added, so the acid is diluted to figure out the exact titrtion point.

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