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vitfil [10]
3 years ago
15

Would anyone be able to give a good explanation of what

Chemistry
1 answer:
mezya [45]3 years ago
3 0

Answer and Explanation:

Aspirin is odorless, but when left exposed to air in the environment, it gradually hydrolyzes into salicylic acid and acetic acid as that is the precursor for synthesizing Aspirin.

Using this hydrolyzed aspirin for titration would not be advised, because it would affect the reading of the titration. Ordinarily, apsirin is a weak acid and direct titration of aspirin is problematic because it hydrolyzes pretty fast to salicylic acid— leading to an unwanted side reaction which may or may not go to completion. Therefore, excess base must be added and heat is supplied to the mixture so that neutralization and hydrolysis are complete. The remaining base is then titrated. This is called back titration.

Now, in back titration, instead of using solution whose concentration is expected to be known, we rather use excess volume of reactant which has been left over after the completion of a reaction with the analyte.

In this case, we use an alkali, preferably NaOH (1.0 mol/dm³). Te unused NaOH remaining after the hydrolysis is titrated against a standard HCl (0.1 mol/dm³). Then from the reaction equation of the aspirin and sodium hydroxide, the amount of NaOH required for the hydrolysis can be calculated.

Answering whether the titration goes up or down, it would be observed that the titration reading would GO DOWN because the exposed aspirin used has experienced some form of hydrolysis before it was used for titration, so the hydrolysis reaction it would undergo with acetyl-salicylic acid would be minimal, and this would affect the titration reading.

But if the aspirin wasn't left exposed to the environment, the reading would go up since more hydrolysis would take place in this case.

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in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc
Margarita [4]

Answer:  0.1161 grams of mercury(II) sulfide) form.

Explanation:

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}     .....(1)

a) Molarity of Na_2S solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol

\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol

b) Molarity of Hg(NO_3)_2 solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol

Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3

According to stoichiometry :

1 mole of Hg(NO_3)_2 reacts with 1 mole of Na_2S

Thus 0.0005 moles of HgNO_3 reacts with=\frac{1}{1}\times 0.0005=0.0005 moles of Hg(NO_3)_2

Thus Hg(NO_3)_2 is the limiting reagent and Na_2S is the excess reagent.

According to stoichiometry :

1 mole of Hg(NO_3)_2 forms=  1 mole of Hg_2S

Thus 0.0005 moles of Hg(NO_3)_2 forms=\frac{1}{1}\times 0.0005=0.0005 moles of Hg_2S

mass of H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g

Thus 0.1161 grams of mercury(II) sulfide) form.

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