# Chemistry (Acids and Bases)

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).

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1. 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
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

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2. thank you very much

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