A solution formed by dissolving an antacid tablet has a PH of 9.18 at 25 c. Calculate [H],[OH] and pOH.
[H] = 1.0 x 10^-9.18
[OH] = 1.0 x 10^-0.82
pOH = 0.82
NOPE.
pH = -log (H^+)
9.18 = -log(H^+) or (H^+) = 10^-9.18. OK here.
pOH + pH = 14 so pOH = 14 - pH = 14 - 9.18 = 4.82 is correct answer.
Then pOH = -log(OH^-) or (OH^-) = 10^-4.82 is correct answer.
To calculate [H+], [OH-], and pOH, we can use the equation for pH and pOH:
pH = -log[H+]
pOH = -log[OH-]
Given that the pH of the solution is 9.18, we can calculate [H+] using the pH equation:
[H+] = 10^(-pH)
[H+] = 10^(-9.18)
[H+] ≈ 6.734 × 10^(-10) M
To calculate [OH-], we can use the fact that in water at 25°C, the concentration of [H+] multiplied by the concentration of [OH-] is equal to 1 x 10^(-14) M^2:
[OH-] = (1 x 10^(-14)) / [H+]
[OH-] = (1 x 10^(-14)) / 6.734 × 10^(-10)
[OH-] ≈ 1.486 × 10^(-5) M
Finally, to calculate pOH, we use the pOH equation:
pOH = -log[OH-]
pOH = -log(1.486 × 10^(-5))
pOH ≈ 4.83
So, the [H+] is approximately 6.734 × 10^(-10) M, [OH-] is approximately 1.486 × 10^(-5) M, and pOH is approximately 4.83.
To calculate the [H+], [OH-], and pOH in a solution with a pH of 9.18, you can use the pH formula:
pH = -log[H+]
First, rearrange the equation to solve for [H+]:
[H+] = 10^(-pH)
Substitute the given pH into the equation:
[H+] = 10^(-9.18)
Using a scientific calculator, calculate [H+]:
[H+] ≈ 6.31 x 10^(-10) M
Since the solution is at a pH of 9.18, which is greater than 7, it is considered a basic solution.
Since the solution is basic, we know that [H+] x [OH-] = 1 x 10^(-14) M^2 (at 25°C).
To find [OH-], rearrange the equation:
[OH-] = (1 x 10^(-14)) / [H+]
Substitute the calculated [H+] into the equation:
[OH-] ≈ (1 x 10^(-14)) / (6.31 x 10^(-10))
Using a scientific calculator, calculate [OH-]:
[OH-] ≈ 1.59 x 10^(-5) M
To find pOH, use the following equation:
pOH = -log[OH-]
Substitute the calculated [OH-] into the equation:
pOH ≈ -log(1.59 x 10^(-5))
Using a scientific calculator, calculate pOH:
pOH ≈ 4.80
Therefore, [H+] ≈ 6.31 x 10^(-10) M, [OH-] ≈ 1.59 x 10^(-5) M, and pOH ≈ 4.80 in the given solution.