Strong electrolytes vs weak electrolytes--some compounds conduct a much stronger current than others.General rules to follow to determine if the compound is a strong or weak electrolyte.
The difference between strong and weak electrolytes is the extent to which the ionic compounds dissociate into ions when placed in water. The greater the amount of dissociation, the greater the electrical conductance of the solution. The strong electrolytes are usually considered to be 100% dissociated, especially in dilute solutions, and weak electrolytes are usually dissociated <10%. The quantitative values of conductance is given for 0.10 M aqueous solutions at 25°C, and is in seimens.cm2/mol.
- The following acids are considered strong acids and are also strong electrolytes: HCl, HBr, HI, HNO3, H2SO4, and HClO4. The rest of the acids are considered weak acids and are also weak electrolytes.
- The hydroxides of Groups I and II are considered strong bases and are also strong electrolytes. The rest of the bases are considered as weak bases and are weak electrolytes.
- Most other ionic compounds are strong electrolytes.
- Halides and cyanides of "heavy metals" are generally weak electrolytes. An example is PbI2.
- Most organic compounds are nonelectrolytes. Exceptions are organic acids and bases.
bond formed between oppositely charged ions because of electrostatic forces of attraction.
- example: sodium + chlorine ----> sodium chloride
a compound composed of ions.
Metal atoms lose electrons to form positive ions, called cations, which are smaller than the original atom.
The loss of an energy level with electrons, plus the excess positive charge draws the remaining electrons toward the nucleus and causes the electron cloud to contract even more.
Nonmetal atoms gain electrons to form negative ions, called anions, which are larger than the original atom.
The addition of an extra electron increases the electron-electron repulsion and causes the electron cloud to expand.
The main Group elements (s and p orbitals) lose or gain electrons to attain a configuration like a noble gas.
Transition elements (d orbitals) lose their s orbital electrons first and then one or more d orbital electron(s).
Ionic bonding can be broken down into 3 steps: ex. sodium chloride
According to Hess's Law, the DH for the formation of one mole of sodium chloride ion-pairs from one mole of sodium atoms and one mole of chlorine atoms is DHrxn = DH1 + DH2 + DH3 = -356 kJ/mol
- Formation of the sodium ions (ionization)
Na(g) + energy -----> Na+(g) (DH = +496 kJ/mol)
- Formation of the chloride ions (electron affinity)
Cl(g) + e- -----> Cl-(g) + energy (DH = -348 kJ/mol)
- Formation of the ion pair (bond energy)
Na+(g) + Cl-(g) -----> NaCl(g) + energy (DH = -504 kJ/mol)
Coulomb's Law is used to calculate the energy in forming the ion-pair. It says that the energy of interaction of two ions is directly proportional to the product of their electrical charges and is inversely proportional to the distance between their centers.
E = kZ1Z2/deq
Z1 & Z2 are charges of the 2 ions
deq = distance between their centers
k = proportionality constant (2.31 x 10-16J pm / ion-pair )
NOTE: All ionic bonding is exothermic.
In the real world ionic compounds don't actually exist as ion-pairs which were formed from naturally occurring gaseous atoms. Instead, they exist as crystalline solids consisting of many ions, each cation being surrounded by anions and each anion being surrounding by cations to form a crystal lattice. Each cation-anion pair represents the release of energy.
Therefore, the reaction of the solid sodium atoms and the diatomic chlorine gas molecule to form the ionic solid sodium chloride can be broken down into 5 steps.
- Vaporize the sodium atoms into gaseous sodium atoms. (use heat of vapaorization)
Na(s) -----> Na(g) (DH = +93 kJ/mol)
- Dissociate the chlorine molecules into chlorine atoms. (use bond energy)
1/2Cl2(g) -----> Cl(g) (DH = +122 kJ/mol)
- Ionize the sodium atoms. This is the oxidation step. (use ionization energy)
Na(g) -----> Na+(g) + e- (DH = +496 kJ/mol)
- Add electron to chlorine atoms to form chloride ions. This is the reduction step. (use electron affinity)
Cl(g) + e- -----> Cl-(g) DH = -348 kJ/mol)
- Formation of the crystal lattice (use lattice energy)
Na+(g) + Cl-(g) -----> NaCl(s) (DH = -780 kJ/mol)
- The overall change in energy for the reaction, DHrxn = -417 kJ/mol
a complete cycle showing all the steps in a process.
Send questions, comments or suggestions to
Gwen Sibert, at the
Roanoke Valley Governor's School
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