1.Topic№1: Chemical equivalent. Significance of solution in vital activity organism. The way of solution concentration expression.

1.Topic№1: Chemical equivalent. Significance of solution in vital activity organism. The way of solution concentration expression.

Basic questions:

1. Introduction. Safety devices, methods of the work in lab.

2. Equivalent of chemical element, Its molar mass

3. Equivalent and molar mass of equivalent of an acid, a base, a salt, oxide.

4. The way of solution concentration expression: percent concentration of the solution, molarity, normality and titer.

5. Significance of solution in vital activity organism.

1) Safety devices, methods of the work in lab

Safe Laboratory Practices

To ensure laboratory safety, follow safe laboratory practices, including the following:

Know about the chemicals and hazards associated with your laboratory.

Know what to do in different emergency situations.

Wear personal protective equipment, as appropriate.

Follow safe practices for working with chemicals.

Ice from a laboratory ice machine should not be used for human consumption.

Do not wear contact lenses around chemicals, fumes, dust particles, or other hazardous materials.

Protect unattended operations from utility failures and other potential problems that could lead to overheating or other hazardous events.

Avoid working alone in a laboratory.

Avoid producing aerosols.

Controlling Laboratory Risks

Administrative have to controls can help minimize laboratory risks. However, safety conscious workers using good laboratory practices are the most important component of laboratory safety.

2) Equivalent of chemical element, Its molar mass

Equivalent

The equivalent of the substance it is its quantity that bonded with one mole of hydrogen atoms or replaces the same quantity of hydrogen atoms in the chemical reaction. For example in the following combinations HC1, H₂S, NH₃, CH₄ the equivalent of Cl, S, N and С is equal to 1 mole,½ mole, 1/3 mole, ¼ mole. A symbol of equivalent is fe and measuring unit is [mole].

3) Equivalent and molar mass of equivalent of an acid, a base, a salt, oxide.

1. Equivalent of an acid

f_e acid = 1/n mole, where n it is acid's basicity

Acid's basicity it is number of hydrogen atoms which possible to replace in the chemical reaction.

f_e H₂SO₄ = ½ mole

2. Equivalent of a base

f_e base = l/n mole, where n is basicity of base or it is number of hydroxyl ions which is possible to replace in the chemical reaction.

f_eAl(OH)₃ = 1/3 mole

3. Equivalent of a salt

f_csalt = 1/n*V mole, where n it is quantity of cation atoms and V it is valence of metal cation in salt.

F_eNa₂SO4 = l/2*1 = ½ mole

4. Equivalent of oxide

f_e oxide = 1/n*V mole, where n it is quantity of cation or anion atoms and V it is valence of cation or anion in salt.

There is one more conception connected with chemical equivalent. It is equivalent weight.

Equivalent weight it is mass of one equivalent expressed in grams.

A symbol of equivalent weight - M_e [gm/mole]

Me (x) = f_e * M(x) - it is formula for equivalent weight calculating.

Me (H₂SO₄) = f_e H₂SO₄ * M(H₂SO₄) = 1/2 * 98 = 49 gm/mole

Equivalent of a substance in the chemical reactions.

It is calculated according to the quantity of elements (atoms) or quantity of

functional groups hydrogen or hydroxyl in the chemical reaction,

H₂SO₄+ 2KOH ---> K₂SO₄ + 2H₂O

f_{e acid} = 1/n₁-n₂, where n₁and n₂ is number of hydrogen atoms in acid before and after the reaction.

H₂SO₄¹/2-1 = 1/1 = l mole.

A1(OH)₃ + H₂SO₄ ---> A1OHSO₄ + 3H₂O

f_{e base} = 1/n₁-n₂, where n₁and n₂ is number of hydroxyl atoms before and after the reaction.

f_e A1(OH)₃ = 1/3-1 = ½ mole.

Chemical equivalent of an element in the combination.

It depends upon valence of a certain element in the compound.

f_e element/x = 1/V mole, where V is element's valence

f_e N/NH₃ = 1/3 mole, f_e N/NO₂ = 1/2mole, f_e N/N₂O = 1 mole,

f_e N/HNO₃ = 1/5 mole.

4)The way of solution concentration expression: percent concentration of the solution, molarity, normality and titer.

Solution is a system consisting of two or more components in different proportions.

Solution = solute + solvent

There are several solutes may be in the solution.

The concentration it is correlation between quantity of solute and solvent.

m solution = m solute + m solvent

1. Percent concentration of the solution.
W% (x)

W% = m solute / m solution * 100% [%]

W% = m solute / m solute + m solvent

Solvent is a component taken in big quantities and maintains constant aggregate state.

Problem:

Calculate concentration of 40 grams of sugar in 360 grams of water.

Solution: W sugar = m sugar/m solution * 100%

W sugar = 40/400 = 0.1 or 10 % solution.

2. Molarity of the solution.

Molarity it is solution concentration expression which shows how many moles of solute is dissolved in 1 liter of the solution.

The symbol of molarity is Cx [mole/1]

Cx = n/V [mole/1], where n - is number of molecules in mole and V is volume of solution.

N = m/M [mole], where M is molecular mass of substance, hence:

С = m/M*V [mole/1], where V is volume of solution in liters.

Problem:

Calculate molarity of 500 ml of NaOH solution containing 20 gm of sodium hydroxide. Solution:

С NaOH = m NaOH /M NaOH*V [mole/1], where V is volume of NaOH solution in liters.

Hence С NaOH = 20/40*0.5 = 1 mole

When we write 2M solution of H₂SO₄ this means that each liter of such solution contains 2 mole of sulfuric acid. C=2mole/1.

3. Normality of the solution.

Normality is concentration that shows the amount of equivalent in one liter if solution.

C_e = n/V [mole/1], where n - is number of equivalent in mole and V is volume of solution.

n_e = m/Me [mole], where M_e is equivalent weight of substance, hence:

C_e= m/M_e*V [mole/1], M_e = f_e * M gm/mole

Problem:

Calculate normality of 500 ml of solution containing 9.8 gm of sulfuric acid. Solution:

C_e H₂SO₄ = m H₂SO₄ /Me H₂SO₄*V [mole/1], M_e = f_e * M gm/mole Me H₂SO₄ = 1/2* 98 = 49 gm/mole

C_e H₂SO₄ = m H₂SO₄ /Me H₂SO₄*V [mole/1] = 9.8/49*0.5 = 0.4 mole/1

4. Titer of the solution.

Titer is amount of grams of solute in one milliliter of solution.

T = m/V [gm/ml]

If titer of the solution is 0.00098 gm/ml this meant that each milliliter of the solution contains 0.00098 grams of solute.

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