If the hydroxide ion concentration in a lake is 10‐10 M what is the lake’s pH?
pOH = -log(OH^-). Solve for pOH.
Then pH + pOH = 14.
Solve for pH.
but when its hydrogen ions, do i use that second equation? ph+poh=14?
The liquid in the stomach has a pH of approximately 2. What is the concentration of hydrogen ions in
this liquid?
for hydroxide ion, there's an additional step to find pH.
after finding pOH, you'd have to plug it into the second equation and solve for pH.
if you were given hydrogen ions, you could find pH in one step.
pH= -log(H^+)
the concentration of hydrogen ions is essentially 10^(-pH)
To determine the lake's pH, we need to know the concentration of the hydrogen ion (H+), which is related to the hydroxide ion (OH-) concentration. In pure water, the concentration of H+ and OH- are equal and can be represented as 1x10^-7 M (at 25°C).
The relationship between the concentration of H+ and OH- is given by the equation: Kw = [H+][OH-] = 1x10^-14
We can rearrange this equation to solve for the concentration of H+:
[H+] = Kw / [OH-]
Substituting the given value of [OH-] as 10^-10 M, we can calculate [H+]:
[H+] = (1x10^-14) / (10^-10)
Simplifying, we get:
[H+] = 10^-14 / 10^-10
When dividing exponential terms with the same base, we subtract their exponents:
[H+] = 10^-4
The pH is defined as the negative logarithm (base 10) of the concentration of H+:
pH = -log[H+]
Substituting the value of [H+], we get:
pH = -log(10^-4)
Using the logarithmic property, we can simplify further:
pH = -(-4) = 4
Therefore, the lake's pH is 4.