Question

Determine the [H+], [OH−] , and pOH of a solution with a pH of 7.41 at 25 °C.

[H+]=

[OH−]=

pOH=

Answer 1

Using the relationship that the sum of pH and pOH is 14.00 at 25 °C, the hydroxide ion concentration [OH−] of a solution with a pH of 7.41 is 2.5 × 10−8 M, the pOH is 6.59, and the hydrogen ion concentration [H+] is 3.9 × 10−9 M.

Step-by-step explanation:

To determine the hydrogen ion concentration [H+], hydroxide ion concentration [OH−], and pOH of a solution with a pH of 7.41 at 25 °C, we can use the relationship that at 25 °C, the sum of the pH and the pOH is always 14.00.

First, to find the pOH:

pOH = 14.00 - pH

pOH = 14.00 - 7.41

pOH = 6.59

Now to find [H+], which can be calculated by:

[H+] = 10−pH

[H+] = 10−7.41 M

[H+] = 3.9 × 10−9 M (approx)

For [OH−], we use:

[OH−] = 10−pOH

[OH−] = 10−6.59 M

[OH−] = 2.5 × 10−8 M (approx)

Answer 2

[H+] = 2.96 x 10^(-9) M

[OH-] = 3.02 x 10^(-7) M

pOH = 6.59

Step-by-Step Explanation:

To determine the [H+], [OH-], and pOH of a solution with a pH of 7.41, we can use the relationship between pH, [H+], and pOH:

pH + pOH = 14

Given that the pH is 7.41, we can subtract it from 14 to find the pOH:

pOH = 14 - pH

pOH = 14 - 7.41

pOH = 6.59

Since pH + pOH = 14, we can also determine the [H+] and [OH-] using the pOH value:

pOH = -log[OH-]

6.59 = -log[OH-]

To solve for [OH-], we can take the antilog of both sides:

[OH-] = 10^(-pOH)

[OH-] = 10^(-6.59)

Using the relationship [H+][OH-] = 1 x 10^(-14) at 25 °C, we can determine [H+]:

[H+] = (1 x 10^(-14)) / [OH-]

[H+] = (1 x 10^(-14)) / (10^(-6.59))

Calculating the values:

[H+] = 2.96 x 10^(-9) M

[OH-] = 3.02 x 10^(-7) M

pOH = 6.59