Answer:
[H₃O⁺] = 2.5 × 10⁻¹⁰ M
pH = 9.6
Explanation:
Step 1: Given data
Concentration of OH⁻ in the solution ([OH⁻]): 4.0 × 10⁻⁵ M
Step 2: Calculate the concentration of H₃O⁺ in the solution
Let's consider the self-ionization of water.
2 H₂O(l) ⇄ H₃O⁺(aq) + OH⁻(aq)
The ion-product of water (Kw) is:
Kw = 1.0 × 10⁻¹⁴ = [H₃O⁺] × [OH⁻]
[H₃O⁺] = 1.0 × 10⁻¹⁴/[OH⁻]
[H₃O⁺] = 1.0 × 10⁻¹⁴/4.0 × 10⁻⁵
[H₃O⁺] = 2.5 × 10⁻¹⁰ M
Step 3: Calculate the pH of the solution
We will use the following expression.
pH = -log [H₃O⁺]
pH = -log 2.5 × 10⁻¹⁰
pH = 9.6
Taking into account the definition of pH and pOH, the pH and [H₃O⁺] of the solution is 9.602 and 2.5×10⁻¹⁰ M respectively.
First of all, pH is a measure of acidity or alkalinity that indicates the amount of hydrogen ions present in a solution or substance.
The pH is defined as the negative base 10 logarithm of the activity of hydrogen ions, that is, the concentration of hydrogen ions or H₃O⁺:
pH= - log [H⁺]= - log [H₃O⁺]
Similarly, pOH is a measure of hydroxyl ions in a solution and is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
The following relationship can be established between pH and pOH:
pH + pOH= 14
In this case, you know that [OH⁻]= 4×10⁻⁵ M. For this concentration, the pOH is calculated as:
pOH= - log (4×10⁻⁵ M)
pOH= 4.398
Then, pH can be calculated as:
pH + 4.398= 14
pH= 14 - 4.398
pH= 9.602
So, the [H₃O⁺] is calculated as:
9.602= - log [H₃O⁺]
[H₃O⁺]= 10⁻⁹ ⁶⁰²
[H₃O⁺]= 2.5×10⁻¹⁰ M
Finally, the pH and [H₃O⁺] of the solution is 9.602 and 2.5×10⁻¹⁰ M respectively.
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