Another question very similar to the one stated above.
Suppose 1.00 x 10-2 mol of KOH and 1.0 x 10-4 mol of HCI are both added to 1.00 L of
water. Calculate the [H3O+] and [OH─] of the resulting solution.
Calculate the hydronium and hydroxide concentration of a solution made by mixing 100 mL 0.020 mol/L NaOH and 25.0 mL of 0.040 mol/L HCl.
I'm having trouble knowing where to start with this question.
3 answers
#1.
mols HCl = M x L = ?
mols NaOH = M x L = ?
Subtract to find excess of which is present.
If excess HCl, that is the mols HCl.
If excess NaOH, that is the mols NaOH.
Then M (of the excess) = mols/total L. That gives either H3O^+ or OH-. The other one is found from
(H3O^+)(OH^-) = Kw = 1E-14.
#2 is done similarly except the problem has calculated mols already.
mols HCl = M x L = ?
mols NaOH = M x L = ?
Subtract to find excess of which is present.
If excess HCl, that is the mols HCl.
If excess NaOH, that is the mols NaOH.
Then M (of the excess) = mols/total L. That gives either H3O^+ or OH-. The other one is found from
(H3O^+)(OH^-) = Kw = 1E-14.
#2 is done similarly except the problem has calculated mols already.
Wow DrBob222, thanks for the clear and concise explanation.
9 answers