THIS HAS TWO PARTS
NaOH(s) was added to 1.0 L of HCl(aq) 0.42 M.
1) Calculate [H3O+] in the solution after the addition of 0.13 mol of NaOH(s).
2) Calculate [H3O+] in the solution after the addition of 0.81 mol of NaOH(s).
These are best done by making an ICE chart.
1L x 0.42M HCl = 0.42 moles HCl
NaOH = 0.13 moles added.
...........HCl + NaOH ==> NaCl + H2O
begin......0.42...0.........0......0
add..............0.13................
react....-0.13..-0.13....+0.13...+0.13
final......0.29....0.......-.13...0.13
You can see that you have 0.29 mole HCl and it's in 1L soln; therefore, HCl being a strong acid (100% ionized), the (H3O^+) = (HCl) = 0.29moles/L = 0.29M.
The second one is done the same way except that the NaOH will be in excess. After you find the (OH^-), convert to (H3O^+) by (H3O^+)(OH^-) = Kw = 1E-14
thank you very much
To solve these problems, we need to understand the reaction between NaOH and HCl and how it affects the concentration of H3O+ in the solution. The reaction is as follows:
NaOH(s) + HCl(aq) → H2O(l) + NaCl(aq)
This is a neutralization reaction, and it involves the reaction of one mole of NaOH with one mole of HCl to form one mole of water and one mole of NaCl.
1) To calculate the concentration of [H3O+] in the solution after the addition of 0.13 mol of NaOH(s), we need to evaluate the stoichiometry of the reaction.
According to the balanced equation, one mole of NaOH reacts with one mole of HCl, producing one mole of H3O+ and one mole of NaCl. Therefore, the amount of HCl reacted with 0.13 mol of NaOH is also 0.13 mol.
The initial concentration of HCl is 0.42 M, which means it originally contains 0.42 mol of HCl in 1.0 L of solution.
When 0.13 mol of NaOH is added, it reacts with 0.13 mol of HCl. This reduces the total amount of HCl to 0.42 mol - 0.13 mol = 0.29 mol.
Since the volume of the solution remains unchanged at 1.0 L, the final concentration is calculated as:
[H3O+] = (0.29 mol) / (1.0 L) = 0.29 M
Therefore, the concentration of [H3O+] in the solution after the addition of 0.13 mol of NaOH(s) is 0.29 M.
2) Similarly, to calculate the concentration of [H3O+] in the solution after the addition of 0.81 mol of NaOH(s), we need to evaluate the stoichiometry of the reaction.
One mole of NaOH reacts with one mole of HCl, so 0.81 mol of NaOH will react with 0.81 mol of HCl.
The initial concentration of HCl is 0.42 M, which means it originally contains 0.42 mol of HCl in 1.0 L of solution.
When 0.81 mol of NaOH is added, it reacts with 0.81 mol of HCl. This reduces the total amount of HCl to 0.42 mol - 0.81 mol = -0.39 mol.
However, it is not physically possible to have a negative amount of HCl. In this case, the reaction is limiting, and all the HCl is consumed. Hence, there will be no H3O+ ions left in the solution.
Therefore, the concentration of [H3O+] in the solution after the addition of 0.81 mol of NaOH(s) is 0 M.