Calculate the hydrogen ion concentration of a 3.55×10-4 M solution of the strong base KOH. Round your answer to 3 significant digits.
pOH = -log(OH^-)
pOH = -log(3.55 x 10^-4) = ??
Then pH + pOH = 14. Convert pOH to pH that way.
To calculate the hydrogen ion concentration of a solution of a strong base, we need to use the concept of the self-ionization of water. Water molecules can undergo a reversible reaction where they donate a proton (H+) to form a hydroxide ion (OH-).
The chemical equation for the self-ionization of water is:
2H2O ⇌ H3O+ + OH-
In pure water, the concentration of H3O+ (which represents the hydrogen ion concentration) is equal to the concentration of OH-. However, when a strong base such as KOH is added to water, it dissociates completely into its constituent ions, which includes hydroxide ions (OH-).
In this case, since KOH is a strong base, its concentration of OH- is equal to the concentration of KOH itself. Therefore, the concentration of OH- is 3.55×10-4 M.
Now, to calculate the hydrogen ion concentration (H+ concentration), we can use the concept of the ion product of water (Kw). In pure water at 25 degrees Celsius, the Kw value equals 1.0 x 10^-14 M^2.
The ion product of water equation is:
Kw = [H3O+][OH-]
Since [H3O+] = [OH-] in pure water, we can substitute the OH- concentration (3.55×10-4 M) into the equation:
Kw = [H3O+][3.55×10-4 M]
Solving for [H3O+], we get:
[H3O+] = Kw / [OH-]
[H3O+] = (1.0 x 10^-14 M^2) / (3.55×10-4 M)
[H3O+] = 2.82 x 10^-11 M
Finally, rounding to 3 significant digits, the hydrogen ion concentration of the 3.55×10-4 M solution of KOH is approximately 2.82 x 10^-11 M.