How can I determine the van't Hoff factor of a substance from its formula?

Question

Owen Ambrose · Accepted Answer

Here's how you do it.

The van't Hoff factor, #i#, is the number of particles formed in a solution from one formula unit of solute. Notice that #i# is a property of the solute. In an ideal solution, #i# does not depend on the concentration of the solution. For a nonelectrolyte If the solute is a nonelectrolyte (i.e. it does not separate into ions in solution), #i = 1# For example, #"sucrose(s) → sucrose (aq)"#. #i = 1#, because 1 molecule of sucrose forms only one particle in solution. For a strong electrolyte If the solute is a strong electrolyte (i.e. it separates into ions in solution), #i > 1#. Some examples are: #"NaCl(s)" → "Na"^+("aq") + "Cl"^"-"("aq"); i = 2# One formula unit of #"NaCl"# will form two particles in solution, an #"Na"^+# ion and a #"Cl"^"-"# ion. #"CaCl"_2(s) → "Ca"^"2+"("aq") + "2Cl"^"-"("aq"); i = 3# One formula unit of #"CaCl"_2# will form three particles in solution, a #"Ca"^"2+"# ion and two #"Cl"^"-"# ions. Here's another example: #"Fe"_2("SO"_4)_3("s") → "2Fe"^"3+"("aq") + "3SO"_4^"2-"("aq"); i = 5# For a weak electrolyte If the solute is a weak electrolyte , it dissociates only to a limited extent. For example, acetic acid is a weak acid. We often set up an ICE table to calculate the number of particles in a 1 mol/L solution. #color(white)(mmmmmm)"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-"# #"I/mol·L"^"-1":color(white)(ml) 1color(white)(mmmmmmml) 0color(white)(mmm) 0# #"C/mol·L"^"-1": color(white)(m)"-"xcolor(white)(mmmmmm) +xcolor(white)(m)+x# #"E/mol·L"^"-1":color(white)(l) 1-xcolor(white)(mmmmmm)xcolor(white)(mmm) x# At equilibrium, we have #1-xcolor(white)(l) "mol of HA", xcolor(white)(l) "mol of H"_3"O"^+, and xcolor(white)(l) "mol of A"^"-"#. #"Total moles" = (1-x + x + x)color(white)(l) "mol" = (1+x)color(white)(l) "mol"#, so #i = 1+x"#. Usually, #x < 0.05#, so #i < 1.05 ≈ 1#.

Jeremiah Adams · Answer

undefined The number of particles that a substance dissociates into during its dissolution in a solution is known as the van't Hoff factor (i). For ionic compounds, the van't Hoff factor is equal to the number of ions that are produced during the dissociation process. For covalent compounds, on the other hand, the van't Hoff factor is typically 1, as they do not dissociate into ions in solution.