Calculate the amount concentration of the cations in the following solutions:
A) 0.5 mol/L sodium carbonate
B) 0.2 mol/L ammonium sulfite
C) 1.5 mol/L iron 3 sulfate
I'll do the middle one.
0.2 mols/L (NH4)2SO3
The cation is NH4^+
For every 1 mol (NH4)2SO3 you have 2 mols NH4^+; therefore, (NH4+) = twice the (NH4)2SO3 and that give 0.2 mols/L x 2 = 0.4 mols/L = 0.4 M.
a is Na2CO3 formula.
c is Fe2(SO4)3 formula.
To calculate the amount concentration of cations in a solution, we need to consider the stoichiometry of the compound and the charge of the cations. Let's go through each solution:
A) 0.5 mol/L sodium carbonate (Na2CO3)
In this compound, the cations are sodium ions (Na+). Since each formula unit of sodium carbonate contains two sodium ions, we multiply the molarity of the solution (0.5 mol/L) by 2 to find the amount concentration of the sodium cations:
0.5 mol/L x 2 = 1 mol/L
Therefore, the amount concentration of sodium cations in a 0.5 mol/L sodium carbonate solution is 1 mol/L.
B) 0.2 mol/L ammonium sulfite (NH4)2SO3
In this compound, the cations are ammonium ions (NH4+). Each formula unit of ammonium sulfite contains two ammonium ions, so we multiply the molarity of the solution (0.2 mol/L) by 2 to find the amount concentration of the ammonium cations:
0.2 mol/L x 2 = 0.4 mol/L
Therefore, the amount concentration of ammonium cations in a 0.2 mol/L ammonium sulfite solution is 0.4 mol/L.
C) 1.5 mol/L iron 3 sulfate (Fe2(SO4)3)
In this compound, the cations are iron(III) ions (Fe3+). Each formula unit of iron(III) sulfate contains two iron(III) ions, so we multiply the molarity of the solution (1.5 mol/L) by 2 to find the amount concentration of the iron(III) cations:
1.5 mol/L x 2 = 3 mol/L
Therefore, the amount concentration of iron(III) cations in a 1.5 mol/L iron(III) sulfate solution is 3 mol/L.