What are [H+] and [OH-] for 0.01 M perchloric acid?
a. If [OH-] in a solution is 10-4 M, what is [H+]?
b. If 1500 mL of water is added to 500 mL of this solution, what will [H+] and [OH-] be?
What are [H+] and [OH-] for 0.01 M perchloric acid?
HClO4 is a strong acid; thus, it is ionized completely and (H^+) = (HClO4). So (H^+) = 0.01 M. Then (H^+) (OH^-) = Ka = 1E-14. You know Ka and you know (H^+), substitute and calculate (OH^-).
a. If [OH-] in a solution is 10-4 M, what is [H+]?
(H^+) (OH^-) = Ka = 1E-14.
b. If 1500 mL of water is added to 500 mL of this solution, what will [H+] and [OH-] be?
Technically you don't know BECAUSE you don't know the final volume BUT if you assume that the volumes are additive (they never are, really, but in this case and many others they are so close that you can't tell the difference. Anyway, assuming they are additive, then use the dilution formula which is as follows:
ml1 x M1 = mL2 x M2
500 x M1 = 1500 x M2 where M1 will be the (H^+) you calculated from the b part. M2 then will be the new concentration of H+. Or (OH^-) is the M1 you calculated from the b part and M2 will be a new calculation for the concentration of OH^-.
There is an easier way to do this. If you take 500 mL of the solution and dilute it to 1500 mL, you know you've dilluted it by a factor of 3 (1500/500 = 3) so you know the solution must be weaker by a factor of 3. Just divice the original H+ or OH- by 3.
Post your work if you run into trouble.
oooops. I made a huge error (two errors). Regarding the b part and the short cut. If you add 1500 mL to the 500 mL, the TOTAL volume will be 2000 mL so
500 x M1 = 2000 x M2 and for the short cut you are diluting the solution by a factor of 4 (2000/500 = 4) so divide by 4 and not 3. Sorry about that. Didn't have my thinking cap on straight.
Thank you @DrBob222 for your responses, I have incorporated your tips into my solution. However, for part b we found the value of [OH-] through the short cut method you described, what happens to the [H+] value does it remain the same as the value we determined in part a, or is there an additional step needed to determine the new value for [H+]?
No. As I read the problem, part a is completely different and is all about 0.01 M HClO4 whereas part b starts with OH = 1E-4 M of some unknown base which makes H^+ in this part b solution 1E-10 M to make (H^+)(OH^-) = 1E-14. So now we add 1500 mL H2O to the 500 mL of the 0.0001 M OH^- .
mL1 x M1 = mL2 x M2
500 mL x 1E-4 M = 2000 mL x M2
M2 = 500 x (10E-5)/2000 = 2.5E-5 M = (OH^-) by the dilution formula.
OR
(1E-4)/4 = (10E-5)/4 = 2.5E-5 M by the shortcut.
Then solve for H^+ from Kw = (H^+)(OH^-) = 1E-14.
(H^+) = 4E-10 M
To find the values of [H+] and [OH-] for a solution, we need to consider the dissociation of the compound. In the case of perchloric acid (HClO4), it is a strong acid and dissociates completely in water, resulting in the release of H+ ions and ClO4- ions.
a. For a 0.01 M perchloric acid solution, the concentration of H+ ions ([H+]) will be equal to the initial concentration of the acid because it dissociates completely. Therefore, [H+] = 0.01 M.
Since the solution contains H+ ions, it is acidic. The concentration of hydroxide ions ([OH-]) can be calculated using the equation Kw = [H+][OH-], where Kw is the ion product of water and is equal to 1 x 10^-14 at 25°C.
We can rearrange the equation to solve for [OH-]:
[OH-] = Kw / [H+]
[OH-] = (1 x 10^-14) / (0.01)
[OH-] = 1 x 10^-12 M
b. When we dilute the solution by adding water, the total volume of the solution increases, but the amount of solute (perchloric acid) remains the same.
To calculate the new concentrations of [H+] and [OH-], we need to consider the dilution formula, which states that the concentration of a solution is inversely proportional to its volume.
Since we are adding 1500 mL of water to 500 mL of the original solution, the total volume becomes 2000 mL (500 mL + 1500 mL).
Using the dilution formula:
[H+]1 x V1 = [H+]2 x V2
(0.01 M) x (500 mL) = [H+]2 x (2000 mL)
[H+]2 = (0.01 M) x (500 mL) / (2000 mL)
[H+]2 = 0.0025 M
Similarly, using the dilution formula for hydroxide ions:
[OH-]1 x V1 = [OH-]2 x V2
(1 x 10^-4 M) x (500 mL) = [OH-]2 x (2000 mL)
[OH-]2 = (1 x 10^-4 M) x (500 mL) / (2000 mL)
[OH-]2 = 2.5 x 10^-5 M
Therefore, after dilution, the new concentration of [H+] is 0.0025 M and the new concentration of [OH-] is 2.5 x 10^-5 M.