H3O+= e^-pH+-error
= e^-pH * e^-+error
=e^-3.72 * e^+-.03
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?
3 answers
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
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
2 answers