We call the cations and anions listed below spectator ions because they do not react with water. When soluble compounds containing one of these spectator ions dissolves in water we can ignore the spectator ion. Soluble compounds of two spectator ions have no direct effect on the pH of water. If you memorize the spectator ions, you can always identify the strong acids, strong bases, and neutral salts. The table list the 12 most common spectator ions, which are easier to remember if you remember the three trios of ions that are grouped together on the periodic table.
|Spectator Cations:||Spectator Anions:|
|Li+||lithium ion||Cl-||chloride ion|
|Na+||sodium ion||Br-||bromide ion|
|K+||potassium ion||I-||iodide ion|
|Rb+||rubidium ion||NO3-||nitrate ion|
|Sr2+||strontium ion||ClO4-||perchlorate ion|
|Ba2+||barium ion||SO42-||sulfate ion|
1. What happens when potassium nitrate is dissolved in water?
KNO3 (s) K+(aq) + NO3-(aq)
Since both K+ and NO3- are spectator ions, KNO3 completely dissolves in water, forming a neutral salt solution.
2. What is the pH of the resulting solution when 0.0500 moles of sulfuric acid, H2SO4, is dissolved in 1.00 L of water?
H2SO4 (l) 2 H+(aq) + SO42-(aq)
SO42- is a spectator ion, so the sulfuric acid dissociates completely. The sulfuric acid releases two moles of H+ for each mole of H2SO4. The pH is:
[H+] = (0.0500 moles H2SO4)(2 moles H+/mol H2SO4) / 1.00 L
[H+] = 0.100 M
pH = -log(H+) = -log(0.100)
pH = 1.000
3. Is an aqueous solution of ammonium nitrate acidic, basic, or neutral?
NH4NO3 (s) NH4+(aq) + NO3-(aq)
Since NO3- is a spectator ion, NH4NO3 completely dissolves in water. The ammonium ion, NH4+, is a weak acid and the solution will be acidic.
4. HF (hydrogen fluoride) is soluble in water. What species are present in the resulting solution when 0.0500 moles of HF are dissolved in water?
HF(l) H+(aq) + F-(aq)
The HF will dissolve in solution. However, since F- is not a spectator ion, the HF will not completely dissociate. There will be HF(aq), H+(aq), and F-(aq) present in solution. The amount of each of these species is determined by the equilibrium constant, which is the topic of this set of practice problems.