If the OH ion concentration in an aqueous solution at 25 celsius is 3.4x10^-3 M the the H concentration in moles per liter in the same solution is?

answers is 2.9x10^-12

i do the -log and i get 2.46 i don't know that else

so i do the -log(3.4x10^-3)=2.46

No, you are making two errors.

#1. The problem asks for (H^+) and gives you the (OH^-). So
(H^+)(OH^-) = Kw = 1E-14
(H^+) = 1E-14/(OH^-) = 1E-14/(3.4E-3) = 2.94E-12 which rounds to 2.9E-12 to two significant figures. That's the answer.
#2. You are using the value of OH as the value for H^+ AND you're changing that to pH which it doesn't ask for.

ok thanks

To find the concentration of H ions (H+) in the solution, you can use the equation for the ion product of water:

Kw = [H+][OH-]

At 25 degrees Celsius, the value of Kw (the ion product constant of water) is known to be 1.0 x 10^-14.

Given that the OH- concentration is 3.4 x 10^-3 M, you can find the H+ concentration as follows:

1. Substitute the OH- concentration into the ion product equation:

1.0 x 10^-14 = [H+][3.4 x 10^-3]

2. Rearrange the equation to solve for [H+]:

[H+] = (1.0 x 10^-14) / (3.4 x 10^-3)

3. Calculate the value:

[H+] = 2.94 x 10^-12 M

Therefore, the H+ concentration in moles per liter in the given aqueous solution is 2.94 x 10^-12 M, which is approximately 2.9 x 10^-12 M.

The problem asks for (H^+) and I see 2.9E-12 on my calculator. You must be punching in the wrong numbers.