what are the concentrations of hydrogen ions and hydroxide ions in a solution that has a oh of 5? *please include work done so I can understand*
pH = 5
pH + pOH = pKw = 14 from which you can find pOH.
Then pH = -log(H^+)
and pOH = -log(OH^-).
If pH = 5, then H^+ = 1E-5. You get that straight from the calculator.
If pH = 5, then pOH = 9 and OH^- = 1E-9 (straight from the calculator) but you can do those in your head.
To determine the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in a solution with a given pH, you need to know the relationship between pH, H+ concentration, and OH- concentration.
The pH of a solution is a measure of its acidity or alkalinity and is defined as the negative logarithm (base 10) of the concentration of H+ ions:
pH = -log [H+]
In neutral water at room temperature, the concentration of H+ ions is equal to the concentration of OH- ions, resulting in a pH of 7. In acidic solutions, the H+ concentration is higher than the OH- concentration, while in alkaline solutions, the OH- concentration is higher than the H+ concentration.
To find the concentrations of H+ and OH- ions, you can start by converting the given OH- concentration to the corresponding pOH, which is defined as:
pOH = -log [OH-]
Given: pOH = 5
To find the concentration of OH- ions ([OH-]) from the pOH, you need to use the inverse log function:
[OH-] = 10^(-pOH)
Now that we have the concentration of OH- ions, we can determine the concentration of H+ ions using the fact that the product of the H+ and OH- concentrations in water is a constant:
[H+] * [OH-] = 1.0 x 10^(-14) (at 25°C)
Since [H+] and [OH-] are equal in neutral water, their concentrations are both (√(1.0 x 10^(-14)), which is approximately 1.0 x 10^(-7) mol/L.
To summarize:
1. Convert pOH to OH- concentration: [OH-] = 10^(-pOH)
For pOH = 5, [OH-] = 10^(-5)
2. Since water is neutral and [H+] = [OH+],
[H+] = √(1.0 x 10^(-14)) ≈ 1.0 x 10^(-7) mol/L
Therefore, the concentrations of hydrogen ions (H+) and hydroxide ions (OH-) in the solution with a pOH of 5 are approximately 1.0 x 10^(-7) mol/L.