What is the pH of a solution with a hydronium ion concentration of 3.5 x 10-4 M? The pOH?

10^-4M

pH + pOH = 14

pH is the negative log of the hydronium ion concentration

pH = -[log(3.5E-4)]

To find the pH and pOH of a solution, you can use the equation:

pH + pOH = 14

Given the hydronium ion concentration of 3.5 x 10^(-4) M, we first need to find the pH using the equation:

pH = -log[H3O+]

pH = -log(3.5 x 10^(-4))

Using a scientific calculator, you can calculate the pH to be approximately 3.46.

Now, to find the pOH, you subtract the pH from 14:

pOH = 14 - pH

pOH = 14 - 3.46

Calculating this, the pOH is approximately 10.54.

To determine the pH and pOH of a solution, you need to understand the relationship between pH, pOH, and the concentration of hydronium ions ([H3O+]) in a solution.

1. pH: The pH of a solution is a measure of its acidity and is defined as the negative logarithm (base 10) of the hydronium ion concentration. The formula to calculate pH is: pH = -log [H3O+].

2. pOH: The pOH of a solution is a measure of its basicity and is defined as the negative logarithm (base 10) of the hydroxide ion concentration ([OH-]). However, the concentration of hydroxide ions in pure water (neutral solution) is very low and can be neglected. Hence, for most common solutions, you can calculate the pOH using the equation: pOH = -log [OH-] ≈ -log [H3O+].

Now, let's use the given hydronium ion concentration ([H3O+]) of 3.5 x 10^-4 M to find the pH and pOH:

pH = -log [H3O+]
pH = -log (3.5 x 10^-4)
pH ≈ -log (3.5) - log (10^-4)
pH ≈ -log (3.5) + 4
Using a scientific calculator, you can calculate -log (3.5) to be approximately 0.455.
pH ≈ 0.455 + 4
pH ≈ 4.455

Thus, the pH of the solution is approximately 4.455.

Since the equation pOH = -log [OH-] ≈ -log [H3O+] can be used for most common solutions, the pOH would also be approximately 4.455 in this case.

Note that pH and pOH are related and that the sum of pH and pOH is always equal to 14 in a neutral solution. However, this equation is not necessary to find the pH and pOH for this question.