i need help with this question. any help would be appreciated.
1) What concentration Of H^+(aq) corresponds to each of the following:(a)pH=3.33,(b)pOH=3.33,(c) pH=6.78,(d)pOH=11.11?
a. pH = 3.33
pH = -log(H^+)
3.33 = -log(H^+)
-3.33 = log(H^+)
antilog -3.33 = antilog log(H^+) or
antilog -3.33 = (H^+).
To find the antilog -3.33, punch in -3.33 on your calculator then push the !0x button. Up will come
4.677 x 10^-4 or 0.0004677
For questions involving pOH, just remember pH + pOH = 14.
Sure! I can help you with that. To determine the concentration of H^+(aq) for each given pH or pOH value, we can use the equations:
pH = -log[H^+(aq)]
pOH = -log[OH^-(aq)]
Let's calculate the concentration for each value step by step:
(a) For pH = 3.33:
First, convert the pH value to a concentration using the pH equation:
[H^+(aq)] = 10^(-pH)
[H^+(aq)] = 10^(-3.33)
[H^+(aq)] ≈ 4.48 x 10^(-4) M
So, the concentration of H^+(aq) for pH = 3.33 is approximately 4.48 x 10^(-4) M.
(b) For pOH = 3.33:
To find the concentration of OH^-(aq), first, convert the pOH value to a concentration using the pOH equation:
[OH^-(aq)] = 10^(-pOH)
[OH^-(aq)] = 10^(-3.33)
[OH^-(aq)] ≈ 4.48 x 10^(-4) M
Since H^+(aq) and OH^-(aq) are in a 1:1 ratio in neutral water, the concentration of H^+(aq) is also approximately 4.48 x 10^(-4) M.
(c) For pH = 6.78:
Using the pH equation:
[H^+(aq)] = 10^(-pH)
[H^+(aq)] = 10^(-6.78)
[H^+(aq)] ≈ 1.72 x 10^(-7) M
Therefore, the concentration of H^+(aq) for pH = 6.78 is approximately 1.72 x 10^(-7) M.
(d) For pOH = 11.11:
To find the concentration of OH^-(aq), first, convert the pOH value to a concentration using the pOH equation:
[OH^-(aq)] = 10^(-pOH)
[OH^-(aq)] = 10^(-11.11)
[OH^-(aq)] ≈ 7.07 x 10^(-12) M
Again, since H^+(aq) and OH^-(aq) are in a 1:1 ratio in neutral water, the concentration of H^+(aq) is also approximately 7.07 x 10^(-12) M.
To recap:
(a) pH = 3.33 corresponds to a concentration of approximately 4.48 x 10^(-4) M H^+(aq).
(b) pOH = 3.33 corresponds to a concentration of approximately 4.48 x 10^(-4) M H^+(aq) as well.
(c) pH = 6.78 corresponds to a concentration of approximately 1.72 x 10^(-7) M H^+(aq).
(d) pOH = 11.11 also corresponds to a concentration of approximately 7.07 x 10^(-12) M H^+(aq).
I hope this helps! If you have any more questions, feel free to ask.