# A buret is used to control and measure the amount of titrant that is added

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analyte of unknown concentration

titrant - AgNO3 - a standard solution - known concentration

titration

By accurately measuring the volume of titrant that is added, the amount of sample can be determined.

A buret is used to control and measure the amount of titrant that is added.

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The endpoint is then the point where sufficient indicator has been converted for detection.

analyte + titrant equivalence point then

This last step does NOT require that all indicator be converted - best if only a small percent need to be reacted to make the change visible.

Note the color change.

A material that is stable in a bottle may not remain that way in solution.

A primary standard solution should:

Have long term stability in your solvent.

The last requirement is often based on the procedure used.

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mol A L solution

mmol A mL solution grams A

Formula Weight A liters of solution

dissolve in water

H2SO4 H3O+ + HSO4-

H3O+ + SO42- dissolution

~99%

~1%

1 M H2SO4 - analytical M [H2SO4] = 0.00 M [HSO4-] = 0.99M [SO42-] = 0.01 M [H3O+] = 1.01 M

molA = weightA (grams) Formula weight (g/mol)

M = molesA

liters solution

gramsA = (liters A)(MA)(Formula WtA) This is all great but lets consider what to do when we actually conduct a titration.

The goal is to determine how much of our analyte is present based on the volume of our titrant.

Assume that A is our titrant and B is our analyte for this general example

They are known to react as follows:

aA + bB products You conduct your titration where the

concentration of A is known and add a known amount of it to your sample.

At the end point, you know MA and the volume of A required - in milliliters.

molesA = liters of A x Molarity of A One typically uses milliliter quantities for a

titration so lets that volume unit mmolesA = mlA MA

Based on our balanced equation, we can determine the moles of B in our sample from:

mmoles mmolesBA

b a = R

=

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mmolesB = mmolesA R = mlA MA R

More commonly, we’ll be interested it the percent of a material in a sample so:

mgB = mlA MA R FWB

% B = 100mgmgB =

sample

mlA MA R FWB 100 mgsample

mlA = 22.12

MA = 0.1200

R = 0.5 ( 1 carbonate / 2 H+ ) FW Na2CO3 = 105.99 g/mol

Sample weight = 0.5000 g

% Na

2CO3 = (22.12 ml)(0.1200M)(.5)(105.99g/mol)(100) 500.0 mgsample

= 28.13 %

(19.80 ml)(0.0500M)(2)(19.00g/mol)(100) 92.5 mg

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Equivalent weight Eq Wt =

Determining the number of equivalents in a mole requires that you know the type of reaction and how the species involved actually combine

In other words - you already know R Formula Weight

# equivalents/mole

If the normality of your titrant and the equivalent weight of your analyte is known, your

calculations are simpler.

NAVA = NBVB = equivalentsB %B =

Realistically, you still need to know R so you can use either type of calculations.

mlA NA eq wtB 100 mg sample

Parts per million and parts per billion

These are extensions of the % system which are used for very dilute solutions

ppm = x 106

ppb = x 109 wt solute

wt solution

wt solute wt solution

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Weight/Volume % =

Mass solute Total Volume x 100

If 5 grams of NaCl is dissolved in water to make 200 ml of solution, what is the concentration?

5 g / 200 ml * 100 = 2.5 wt/v%

Saline is a 0.9 wt/v% solution of NaCl in water.

use g and ml

Volume/Volume % =

Volume Solute Total Volume x 100

If 10 ml of alcohol is dissolved in water to make 200 ml of solution, what is the concentration?

10 ml / 200 ml * 100 = 5 V/V%

Alcohol in wine is measured as a V/V%.

Use the same units for both

Weight/Weight % = Mass Solute Total Mass x 100

If a ham contained 5 grams of fat in 200 g of ham, what is the % wt/wt?

5 g / 200g * 100 = 2.5 wt/v%

On the label, it would say 97.5 % fat free.

Use the same units for both

Establishes the relationship between volume of titrant and amount of analyte present.

Most commonly titer is in units of mg analyte / ml titrant.

This system was developed to assist in doing routine calculations. It reduces the amount of time and training for technicians.

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We can determine the number of mg sodium carbonate / ml of HCl by:

mg Na

2CO3 = (1.00 ml HCl)(0.1200M)(.5)(105.99g/mol)

= 6.36 mg

Our titer then is 6.36 mg Na2CO3 / ml HCl.

Calculation of the % sodium carbonate is now reduced to:

% Na2CO3 = 100 x mlHCl x titer / wt sample

This is a very straight forward calculation that can readily be taught to an assistant will minimal training.

A 1.6732 gram sample is dissolved and titrated with HCl (titer of 5.00 mg/ml). 34.60 ml of HCl is required to reach the methyl orange

endpoint.

Determine % Na2CO3 %Na2CO3 =

= 10.31%

(34.50ml)(5.00mg/ml) (100) 1673.2 mg sample

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