Balance the equation in aqueous acidic solution:
IO-(aq) + I-(aq) → I2(s)
i know you have to add either water or H+ but i cant seem to balance it out
thanks for the help!!
2H+ + IO- + I- >>I2 + H2O
thanks!..for a basic solution would you just add OH then?
for example:
Mn2+(aq) + ClO3-(aq) → MnO2(s) + ClO2(g)
To balance the equation IO-(aq) + I-(aq) → I2(s) in aqueous acidic solution, you need to follow a step-by-step approach. Here's how you can do it:
Step 1: Start by balancing elements other than oxygen and hydrogen.
In this equation, iodine (I) is unbalanced on the right side. There is one iodine on the left side and two iodines on the right side, so we can start by adding a coefficient of 2 in front of IO-(aq). The equation now becomes:
2IO-(aq) + I-(aq) → I2(s)
Step 2: Balance the oxygen atoms by adding water (H2O) molecules.
Since this is an acidic solution, you can add water (H2O) molecules to balance the oxygens. On the left side, there are no oxygens, and on the right side, there are two oxygens in I2. Therefore, you can add a coefficient of 2 in front of H2O on the left side of the equation. The equation becomes:
2IO-(aq) + I-(aq) + 2H2O → I2(s)
Step 3: Balance the hydrogen atoms by adding hydrogen ions (H+).
Add the appropriate number of hydrogen ions (H+) to balance the hydrogens. On the right side, there are two hydrogen ions. Therefore, you can add a coefficient of 2 in front of H+ on the left side of the equation. The equation now becomes:
2IO-(aq) + I-(aq) + 2H2O → I2(s) + 2H+(aq)
Step 4: Check and balance charges.
In this case, the charges are already balanced since iodine has a charge of -1 both before and after the reaction.
The balanced equation in acidic solution is:
2IO-(aq) + I-(aq) + 2H2O → I2(s) + 2H+(aq)
Keep in mind that additional steps might be necessary if the reaction were to be balanced in a different solution (basic solution), where hydroxide ions (OH-) would be added instead of hydrogen ions (H+).