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AleksAgata [21]
3 years ago
10

What are metal complex formation titrations, what is an example of their use?​

Chemistry
1 answer:
meriva3 years ago
6 0

Titrations can be classified by the type of reaction. Different types of titration reaction include:

ACID-BASE TITRATIONS are based on the neutralization reaction between the analyte and an acidic or basic titrant. These most commonly use a pH indicator, a pH meter, or a conductance meter to determine the endpoint.

REDOX TITRATIONS are based on an oxidation-reduction reaction between the analyte and titrant. These most commonly use a potentiometer or a redox indicator to determine the endpoint. Frequently either the reactants or the titrant have a colour intense enough that an additional indicator is not needed.

COMPLEXOMETRIC TITRATIONS are based on the formation of a complex between the analyte and the titrant. The chelating agent EDTA is very commonly used to titrate metal ions in solution. These titrations generally require specialized indicators that form weaker complexes with the analyte. A common example is Eriochrome Black T for the titration of calcium and magnesium ions.

A form of titration can also be used to determine the concentration of a virus or bacterium. The original sample is diluted (in some fixed ratio, such as 1:1, 1:2, 1:4, 1:8, etc.) until the last dilution does not give a positive test for the presence of the virus. This value, the titre, may be based on TCID50, EID50, ELD50, LD50 or pfu. This procedure is more commonly known as an assay.

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Answer:

The new volume will be 5.80 L

Explanation:

Step 1: Data given

The initial volume = 29.0 L

The initial pressure = 562 torr = 562 /760 atm = 0.739474 atm

The pressure changes to 2810 torr = 2810 /760 atm = 3.697368 atm

The temperature and number of moles will not change

Step 2: Calculate the new volume

P1*V1 = P2*V2

⇒With P1 = the initial pressure = 0.739474 atm

⇒with V1 = the initial volume = 29.0 L

⇒with P2 = the increased pressure = 3.697368 atm

⇒with V2 = The new volume = TO BE DETERMINED

0.739474 atm * 29.0 L = 3.697368 atm * V2

V2 = 5.80 L

OR

562 torr * 29.0 L = 2810 torr * V2

V2 = 5.80 L

The new volume will be 5.80 L

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Explanation:

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In the basin, the sediments are accumulated in their final resting position.

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