The question:

The pH of a solution is defined as
pH = -log[H3O+].
If the pH of a solution is 3.72(+/- 0.03), what is the [H3O+] and its absolute standard deviation?

Answer: How would I carry this out? I know to find the hydronium concentration wouldn't i just take e^3.72? How do I do it the way they are asking for?

H3O+= e^-pH+-error

= e^-pH * e^-+error
=e^-3.72 * e^+-.03

pH = -log(H3O^+)

3.72(+/-0.03) = -log(H3O^+)
So the answer is antilog -3.72 (+/- ?) = 0.0001905 +/- ?.
The formula for handling antilogs (10^-pH) is (s/y)= 2.303 x (abs std dev) = (s/y) = 2.303 x (0.03)
(s/y) = 0.0691
s = 0.0691 x 0.0001905 = 0.000013
answer is
0.0001905 +/-0.000013 which I would round to 0.00019 +/-0.00001
Check my work, especially for typos

I believe (H3O^+) = 10^-pH

To find the hydronium ion concentration ([H3O+]) from the given pH value, you are correct that you can use the formula e^pH. However, since the given pH has an associated uncertainty or uncertainty range (+/- 0.03), we need to consider this uncertainty when calculating [H3O+] and its absolute standard deviation.

To calculate [H3O+], you can use the formula [H3O+] = e^pH. In this case, [H3O+] = e^3.72. Evaluating this expression will give you the concentration.

To calculate the absolute standard deviation, we need to consider the uncertainty in the pH value. The absolute standard deviation can be calculated using the formula:
Absolute Standard Deviation = e^(pH + ΔpH) - e^(pH - ΔpH)

In this case, ΔpH is the uncertainty range, which is +/- 0.03. So, the absolute standard deviation = e^(3.72 + 0.03) - e^(3.72 - 0.03).

By evaluating these expressions, you will get the [H3O+] concentration and its absolute standard deviation, in accordance with the given pH value and its uncertainty range.