Asked by Hal
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.
I'm having trouble knowing where to start with this question.
Answers
Answered by
Hal
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.
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.
Answered by
DrBob222
#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.
Answered by
Hal
Wow DrBob222, thanks for the clear and concise explanation.
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