NaOH was added to 1.0 L of HI(aq) 0.40 M. Calculate H3O in the solution after the addition of .14 mol of NaOH.
Calculate H3O in the solution after the addition of 0.91 mol of NaOH.
To calculate the concentration of \(H_3O^+\) in the solution after the addition of NaOH, we need to use the stoichiometry of the reaction between HI and NaOH.
The balanced chemical equation for the reaction between HI and NaOH is:
\(HI + NaOH \rightarrow NaI + H_2O\)
From the balanced equation, we can see that the stoichiometric ratio between HI and NaOH is 1:1. This means that 1 mole of HI reacts with 1 mole of NaOH.
1. Calculation for the addition of 0.14 mol of NaOH:
Given:
- Initial concentration of HI (HIaq) = 0.40 M
- Volume of solution = 1.0 L
- Moles of NaOH added = 0.14 mol
Since the stoichiometric ratio between HI and NaOH is 1:1, the 0.14 mol of NaOH will react with an equal number of moles of HI.
Therefore, the moles of HI that react with 0.14 mol of NaOH will also be 0.14 mol.
To calculate the concentration of HI after the reaction, we need to find the moles of HI remaining and divide it by the volume of the solution.
Moles of HI remaining = Initial moles of HI - Moles of HI that reacted
Initial moles of HI = Initial concentration of HI x Volume of solution
= 0.40 M x 1.0 L
= 0.40 mol
Moles of HI remaining = 0.40 mol - 0.14 mol
= 0.26 mol
Concentration of HI after the reaction = Moles of HI remaining / Volume of solution
= 0.26 mol / 1.0 L
= 0.26 M
Since this reaction is a strong acid-strong base reaction, we can assume complete dissociation of HI. This means that all the HI molecules ionize to produce \(H_3O^+\) ions.
Therefore, the concentration of \(H_3O^+\) in the solution is equal to the concentration of HI remaining after the reaction, which is 0.26 M.
Hence, the concentration of \(H_3O^+\) in the solution after the addition of 0.14 mol of NaOH is 0.26 M.
2. Calculation for the addition of 0.91 mol of NaOH:
Using the same approach as above, we can calculate the concentration of \(H_3O^+\) in the solution after the addition of 0.91 mol of NaOH.
Given:
- Initial concentration of HI (HIaq) = 0.40 M
- Volume of solution = 1.0 L
- Moles of NaOH added = 0.91 mol
The moles of HI that react with 0.91 mol of NaOH will also be 0.91 mol. Therefore, the moles of HI remaining after the reaction will be:
Moles of HI remaining = Initial moles of HI - Moles of HI that reacted
= 0.40 mol - 0.91 mol
= -0.51 mol
We have a negative value for the moles of HI remaining. This indicates that all the HI molecules have reacted, and we have an excess of NaOH.
Since all the HI molecules have reacted, we have no more \(H_3O^+\) ions in the solution.
Therefore, the concentration of \(H_3O^+\) in the solution after the addition of 0.91 mol of NaOH is 0 M.
In summary:
- After the addition of 0.14 mol of NaOH, the concentration of \(H_3O^+\) is 0.26 M.
- After the addition of 0.91 mol of NaOH, the concentration of \(H_3O^+\) is 0 M.