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2 years ago

What are the difficulties encountered in precipitation titration

1 answer:

8 0

One difficulty encountered in precipitation titration is that it is hard to determine the exact end point of its reaction.

Precipitation titration is a titration in which a reaction occurs from the analyte and titrant to form an insoluble precipitate.

With the use of silver for the titrations, (argentometric) we are able to develop many precipitation reactions.

The precipitation titrimetry methods with the use of argentometry includes

• Mohr’s Method

• Fajan’s Method

• Volhard’s Method

Difficulties encountered in precipitation titration includes

Getting the exact end point is hard.

it is a very slow titration method.

it includes periods of filtration and cooling thereby reducing the reactions available for this type of titration.

See more on Precipitation: brainly.com/question/20628792

You might be interested in

Calculate the mass percent (m/m) of a solution prepared by dissolving 51.56 g of NaCl in 164.2 g of H2O. Express your answer to

denis-greek [22]

Answer:

"23.896%" is the right answer.

Explanation:

The given values are:

Mass of NaCl,

= 51.56 g

Mass of H₂O,

= 165.6 g

As we know,

⇒ Mass of solution = $Mass \ of \ (NaCl+H_2O)$

= $51.56+164.2$

= $215.76 \ g$

hence,

⇒ $Mass \ percent =\frac{Mass \ of \ NaCl}{Mass \ of \ solution}\times 100$

$=\frac{51.56}{215.76}\times 100$

$=23.896 \ percent$

4 0

3 years ago

How does the solubility of a gas change with decreasing temperature? A. It increases. B. It decreases. C. It is uniform. D. Temp

vitfil [10]

A it increases is the correct answer.

Good luck and I hope this helps.!

3 0

3 years ago

what is the molarity of a solution made by dissolving 18.5 grams of NaCL into enough water to reach 450.0ML

ValentinkaMS [17]

Answer:

0.702M

Explanation:

Molarity of a solution, which refers to the molar concentration of that solution can be calculated thus;

Molarity = number of moles (n) ÷ volume (V)

Firstly, we convert 18.5 grams of NaCl to moles using the formula; mole = mass/molar mass

molar mass of NaCl = 23 + 35.5 = 58.5g/mol

mole = 18.5/58.5

mole = 0.32moles

Volume of water (V) = 450mL = 450/1000 = 0.450L

Molarity = n/V

Molarity = 0.32/0.450

Molarity = 0.702M

4 0

3 years ago

Which molecular solid would have the highest melting point?

TEA [102]

Answer:

Choice B. The solid with hydrogen bonding.

Assumption: the molecules in the four choices are of similar sizes.

Explanation:

Molecules in a molecular solid are held intact with intermolecular forces. To melt the solid, it is necessary to overcome these forces. The stronger the intermolecular forces, the more energy will be required to overcome these attractions and melt the solid. That corresponds to a high melting point.

For molecules of similar sizes,

The strength of hydrogen bonding will be stronger than the strength of dipole-dipole attractions.
The strength of dipole-dipole attractions (also known as permanent dipole) will be stronger than the strength of the induced dipole attractions (also known as London Dispersion Forces.)

That is:

Strength of Hydrogen bond > Strength of Dipole-dipole attractions > Strength of Induced dipole attractions.

Accordingly,

Melting point due to Hydrogen bond > Melting point due to Dipole-dipole attractions > Melting point due to Induced Dipole attractions.

Induced dipole is possible between all molecules.
Dipole-dipole force is possible only between polar molecules.
Hydrogen bonds are possible only in molecules that contain $\rm H$ atoms that are bonded directly to atoms of $\rm F$ , $\rm O$ , or $\rm N$ .

As a result, induced dipoles are the only force possible between molecules of the solid in choice C. Assume that the molecules are of similar sizes, such that the strengths of induced dipole are similar for these molecules.

Melting point in choice B > Melting point in choice D > Melting point in choice A and C.

8 0

3 years ago

Which element is larger lithium or bromine

Lera25 [3.4K]

Lithium i think!!
hope this helps

5 0

2 years ago