1 equivalent of acid with 1 equivalent of base, 1 equivalent of oxidant with 1 equivalent of reductant, irrespective of the respective molar concentrations and the nature of the antagonistic substances. The concept of equivalent-gram is thus related to the exchange of active particles (protons or electrons) between such substances.
## II.1.5.1. Définition de l’équivalent-gramme
- **Equivalent-gram (meq)** is defined as the amount (in grams) of a substance that engages one mole of active particles. To determine the equivalent weight, the following steps must be taken:
- Identify the nature of the substance,
- Write its dissociation reaction,
- Highlight the number of active particles involved per mole of substance (denoted as $$p$$). Thus:
- Formula: $$m_{eq} = rac{M_{olar}}{p}$$
- Where:
- $$m_{eq}$$ = equivalent-gram mass,
- $$M_{olar}$$ = molar mass,
- $$p$$ = number of active particles and expressed in eq·mol⁻¹.
## II.1.5.2. Détermination du nombre de particules actives p
- The number of active particles $$p$$ involved per mole of substance is determined for different cases:
- **For salts:** $$p$$ equals the total number of released positive or negative charges by 1 mole of salt.
- Example: For $$NaCl
ightarrow Na^+ + Cl^-$$; $$p = 1 eq/mol$$ and $$m_{eq} = rac{M}{1} = 58.5 ext{ g/eq}$$
- For $$BaCl_2
ightarrow Ba^{2+} + 2 Cl^{-}$$; $$p = 2 eq/mol$$ and $$m_{eq} = rac{M}{2} = rac{208.23}{2} = 104.115 ext{ g/eq}$$
- **For acids and bases:** $$p$$ equals the number of protons (H^+ or H3O^+) released by 1 mole of acid or captured by 1 mole of base. It can also be defined by the number of OH⁻ ions, since in aqueous solution, each H⁺ ion is accompanied by as many OH⁻ ions.
- **Examples:**
- For acids:
- $$HCl + H_2O
ightarrow H_3O^+ + Cl^-$$; $$p = 1 eq/mol$$ and $$m_{eq} = rac{M}{1} = 36.5 ext{ g/eq}$$
- $$H_2SO_4 + 2H_2O
ightarrow 2H_3O^+ + SO_4^{2-}$$; $$p = 2 eq/mol$$ and $$m_{eq} = rac{M}{2} = rac{98.079}{2} = 49 ext{ g/eq}$$
- For bases:
- $$NaOH
ightarrow Na^+ + OH^-$$; $$p = 1 eq/mol$$ and $$m_{eq} = rac{M}{1} = 40 ext{ g/eq}$$
- $$Ba(OH)_2
ightarrow Ba^{2+} + 2 OH^-$$; $$p = 2 eq/mol$$ and $$m_{eq} = rac{M}{2} = rac{171.34}{2} = 85.7 ext{ g/eq}$$
- **For oxidants and reductors:** $$p$$ equals the number of electrons released by 1 mole of reductant or captured by 1 mole of oxidant.
- **Examples:**
- For oxidants:
- $$KMnO_4 + 8H^+ + 5e^{-}
ightarrow K^{+} + Mn^{2+} + 4H_2O$$ ; $$p = 5 eq/mol$$ and $$m_{eq} = rac{M}{5} = rac{158}{5} = 31.6 ext{ g/eq}$$
- For reductors:
- $$2 Na_2S_2O_3
ightarrow 4 Na^+ + S_4O_6^{2-} + 2 e^{-}$$ ; $$p = 1 eq/mol$$ and $$m_{eq} = rac{M}{1} = 158 ext{ g/eq}$$
- $$H_2O_2
ightarrow 2H^+ + O_2 + 2 e^{-}$$ ; $$p = 2 eq/mol$$ and $$m_{eq} = rac{M}{2} = rac{34.01}{2} = 17 ext{ g/eq}$$
- Thus: $$n_{eq} = p imes n$$.
# II.1.6. Relation entre les concentrations Cm, CM et Céq
- Knowledge of one concentration allows us to deduce all the other concentrations. The following relationships can be established:
$$C_{mA} = rac{m_A}{V_A} = rac{M imes n_A}{V_A} = M imes C_{mA}$$
- Equivalent concentration relationship is expressed as:
$$C_{eqA} = rac{n_{eq}}{V_A} = p imes rac{n_A}{V_A} = p imes C_{mA}$$
- Thus,
$$C_{MA} = rac{C_{eqA}}{p} = rac{C_{mA} imes M}{M}
# II.2. Composition rapportée à l’unité de masse (m) ou au nombre de mole (n) (titre)
## II.2.1. Fraction ou titre massique
- Represents the ratio of the mass of a constituent $$i$$, contained within a certain volume of solution to the mass of that volume of solution.
- Noted as $$ au_i$$ or $$x_i$$ with $$x_i = au_i = rac{m_i}{ extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle extstyle { ext{sum of individual masses}}} = rac{m_i}{m_s}$$
- **Important Notes:**
- This ratio is dimensionless and must be between 0 and 1.
- The sum of the mass fractions of all constituents equals 1 ($$ extstyle{ extstyle extstyle extstyle{ extstyle extstyle extstyle extstyle{ extstyle extstyle extstyle extstyle{ extstyle extstyle extstyle extstyle{ extstyle extstyle extstyle extstyle (x_i = 1)}}}}}}$$
## II.2.2. Le pourcentage en masse
- Commonly expressed as a percentage to handle numbers between 0 and 100.
- For example, a solution of constituent $$i$$ at 10% contains 10g of compound $$i$$ for every 100g of solution. Thus, the percent mass calculation:
$$ ext{%} i = rac{100 m_i}{ extstyle{ ext{{sum of individual masses}}}}=100 imes au_i = 100 imes x_i$$
- The percent mass is particularly relevant for very concentrated solutions.
## II.2.3. Fraction molaire
- Represents the ratio of the number of moles of the constituent $$i$$ contained within a certain volume of solution to the total number of moles of all constituents present in that volume of solution.
- A mole fraction $$X_i$$ is a dimensionless number calculated by the following relationship:
$$X_i = rac{n_i}{ extstyle{ ext{sum of moles}}} = rac{n_i}{n_t}$$
- Mole fractions are rarely used to express solute concentrations in dilute solutions; however, they serve to express the composition of mixtures.
## II.2.4. Molalité
- Refers to the number of moles of solute per 1 kg of solvent.
- Noted as $$ ext{y}$$ and expressed in mol/kg of solvent (or molal).
- Formula: $$ ext{y} = rac{n}{m_e(kg)}$$
- Note: Rarely used, but occasionally referred to as molal concentration despite being expressed in terms of mass rather than volume.
## II.2.5. Relation entre les titres : $$ ext{y}, x_A, X_A$$
- By expressing masses relative to the mass fraction and the mass of the solution $$m_s$$, we obtain:
$$ ext{y}_A = rac{n_A}{m_e(kg)} = 1000 rac{n_A}{m_e} = 1000 rac{m_A}{M imes x_emS} = 1000 rac{x_A}{M(1 - x_A)}$$
- When using the mole fraction $$X_A$$ and the molar mass of the solvent $$M_e$$, we can establish:
$$ ext{y}_A = 1000 rac{X_A}{M_e(1 - X_A)}$$
# II.3. Relation entre les concentrations et les titres
- To convert between titles and concentrations, the density or mass density must be known. Consequently, the following relationships can be established:
## II.3.1. Relation entre la fraction massique et les concentrations
- $$C_A = rac{n_A}{V} = rac{m_A}{M imes V} = rac{x imes m_s}{M imes V} = rac{x imes
ho_A imes M}{M}$$
- $$C_{eqA} = p imes rac{x imes
ho_A}{M}$$
- $$C_{mA} = x imes
ho_A$$
## II.3.2. Relation entre la molalité et les concentrations
- $$ ext{y}_A = rac{n_A}{m_e(kg)} = 1000 rac{n_A}{m_e} = 1000 rac{m_A}{M ext{s}_A}$$
- $$ ext{y}_A = 1000 rac{C_{mA}}{M(
ho_A - C_{mA})} = 1000 rac{C_{eqA}}{M(
ho_A - C_{eqA})}$$
- Knowing the mass density allows for the retrieval of all other quantitative values of a solution from a single dimension of series.
# III. AUTRES EXPRESSIONS DE LA COMPOSITION D’UNE SOLUTION
## III.1. Titre ou Degré Chlorométrique d’une solution d’eau de javel (NaClO)
- It is the volume or number of liters of Cl2 produced by 1 liter of a solution of sodium hypochlorite or bleach (NaClO) under normal conditions according to the equation:
$$NaClO + 2 HCl
ightarrow NaCl + H_2O + Cl_2$$
- Noted as $$D$$ and expressed in degrees:
$$D = rac{22.4 imes C_{NaClO}}{liter} = rac{11.2 imes C_{eq} NaClO}{liter}$$
- *Exercise 1:* Determine both molar and equivalent concentrations of \[AC} NaClO with a chlorometric degree of 0.56.
- Additionally, when this solution is diluted by one-tenth, calculate the new chlorometric degree.
## III.2. Titre Volumique d’une solution d’eau oxygénée (H2O2)
- It is the volume or the number of liters of O2 produced by 1 liter of a solution of hydrogen peroxide under normal conditions according to the equation:
$$H_2O_2
ightarrow H_2O + rac{1}{2} O_2$$
- Noted as $$TV$$ and expressed in volume:
$$TV = 11.2 imes C_{H_2O_2} = 5.6 imes C_{eq} H_2O_2$$
- *Exercise 2:* Calculate the volume of hydrogen peroxide of volume title $$TV = 0.56$$ required to neutralize 10 mL of a solution with a concentration of 15.8 g.L⁻¹ of KMnO4.
# IV. APPLICATIONS : Cas des solutions binaires
- A binary solution is characterized by containing a solvent and a single solute. Let’s denote it as:
- $$m_s A$$: mass of solution A,
- $$m_A$$: mass of solute,
- $$m_e$$: mass of water,
- $$X_A$$: mole fraction of A,
- $$x_A$$: mass fraction of solute A.
## IV.1. Notion de dilution
- **Dilution** is the process of decreasing the concentration of solution A by adding water to create diluted solution B.
- Schematically:
$$ ext{Solution A { } V_A C_{MA}
ho_A; Céq; C_{mA} + V_{eaj}}$$
$$ ext{Solution B { } V_B C_{MB}
ho_B; Céq; C_{mB}}$$
## Laws of dilution:
1. **Conservation of mass:**
- $$n_A = n_B$$
- hence,
- $$C_{MA} imes V_A =C_{MB} imes V_B$$,
- and,
- $$m_A = m_B$$
- hence,
$$C_{mA} imes V_A =C_{mB} imes V_B$$,
- for equivalent concentrations:
- $$n_{eqA} = n_{eqB}$$
- hence,
$$C_{eqA} imes V_A =C_{eqB} imes V_B$$
2. **Addition effect**:
- $$m_s B = m_s A + m_{eaj}$$
- and,
- $$V_B = V_A + V_{eaj}$$
- **Remarks:**
1. During dilution, all quantities related to the volume of solution, mass of solvent, and solution change.
2. The concentrated solution from which the new solution is prepared is referred to as the “mother solution,” while the new, less concentrated solution is termed the diluent solution.
*Exercise 3:* A solution B obtained by diluting a 50% commercial KOH solution (M(KOH) = 56.1 g/mol) by 10 times. Calculate the molar concentration and density of the diluted solution B.
# IV.2. Dosage ou neutralisation et équivalence ou titrage
- This process constitutes mixing two antagonistic solutions A and B (acid + base or oxidant + reductant). A total reaction occurs between the two solutions. At the end of the reaction (equivalence point), two outcomes may occur:
1. If both solutions have the same number of equivalents, they are said to be in total neutralization:
- $$n_{eqA} = n_{eqB}
ightarrow C_{eqA}V_A = C_{eqB}V_B$$
2. If one solution has more equivalents than the other, it is termed partial neutralization.
- **Remark:** To perform a titration, A and B must be antagonistic, and the reaction must meet the following conditions:
- Fast and spontaneous reaction without a catalyst or temperature elevation,
- Total reaction signaled by a change of color or physical state of the solution (such as the formation of a precipitate).
- Caution: At equivalence, one should never use $$n_A = n_B$$, which is valid only when the antagonistic substances exhibit the same number of active particles.
*Exercise 4:* A diluted solution B of KOH (M(KOH) = 56.1 g/mol) has a concentration of 1.3 eq/L. Calculate the volume of B needed to neutralize:
1. 10 mL of 1.5 M HCl.
2. 12 mL of a 10% solution of H2SO4 with a density of 1.2.
3. 27.5g of 0.8 molal HClO4, where M(HClO4) = 100.5 g/mol.
4. 50 g of HNO3 solution with a mole fraction of 0.01, where M(HNO3) = 63 g/mol.
# IV.3. Mélange de solutions de même nature
- Let solution C be formed by mixing two aqueous solutions A & B of the same nature.
A:
- $$V_A C_{MA}
ho_A ext{ y}_A m_{SA} m_A me_A x_A; C_{eqA} C_{mA}$$
B:
- $$V_B C_{MB}
ho_B ext{ y}_B m_{SB} m_B me_B x_B; C_{eqB C_{mB}$$
- Solution C has:
- $$V_C C_{MC}
ho_C ext{ y}_C m_{SC} m_C me_C x_C; C_{eqC C_{mC}$$
- The mixing is governed by the following relationships:
- $$V_C = V_A + V_B$$
- $$m_{SC} =
ho_C V_C = m_{SA} + m_{SB} =
ho_A V_A +
ho_B V_B$$
- $$m_C = m_A + m_B = x_C m_{SC} = x_C(m_{SA} + m_{SB}) = x_C(
ho_A V_A +
ho_B V_B)$$
- $$m_{eC} = m_{eA} + m_{eB} = x_{eC} m_{SC} = x_{eC}(m_{SA} + m_{SB}) = (1-x_C)(
ho_A V_A +
ho_B V_B) = (1-x_A)
ho_A V_A + (1-x_B)
ho_B V_B $$
- *Exercise 5:* On mixing 250 mL of a 40% Na2CO3 solution (density = 1.824) with 525 g of a Na2CO3 solution with an equivalent concentration of 16.17 eq/L and a density of 1.775 g/mL, calculate:
1. The molar concentrations of solutions A and B.
2. The molar concentration of the final solution, marked SC.
3. The molality of the final solution.
**Note:** The course is overseen by Pr Abdou Karim Diagne DIAW, focusing on the Faculty of Sciences and Techniques at the Université Cheikh Anta DIOP de Dakar, providing a comprehensive exhibition on solution composition measures and their implications in chemistry studies and applications.
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**Responsabile du cours:**
Pr Abdou Karim Diagne DIAW
Département de Chimie
FACULTÉ DES SCIENCES ET TECHNIQUES
Université Cheikh Anta DIOP de Dakar
Each main point in the notes is separated into organized hierarchical structures that provide a comprehensive overview. All necessary definitions, explanations, examples, and mathematical functions are documented precisely for thorough understanding of solution composition. The document serves as an educational report summarizing significant chemical principles applicable in laboratory and practical scenarios, forming a basis for further study in chemistry disciplines.
*Note: For full elaboration, individual calculations related to exercises can be appended as required.*
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