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


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.


stoichiometric addition

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.

React rapidly with your analyte React completely with your analyte Be selective for your analyte

The last requirement is often based on the procedure used.


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



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



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 =


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


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


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.


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


Determine % Na2CO3 %Na2CO3 =

= 10.31%

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




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