12.85
1.70
The concentration of hydroxide ions (OH⁻) in a solution can be represented as [OH⁻]. This concentration can be determined using the formula: [OH⁻] = Kw / [H⁺], where Kw is the ion product of water (1.0 x 10⁻¹⁴ at 25°C) and [H⁺] is the concentration of hydrogen ions in the solution. Alternatively, in basic solutions, [OH⁻] can be calculated directly from the pOH using the relation [OH⁻] = 10^(-pOH).
The statement "H⁺ plus OH⁻ equals 14" is a misunderstanding of the pH scale. In pure water at 25°C, the concentration of hydrogen ions (H⁺) and hydroxide ions (OH⁻) is equal, and their product results in a constant (Kw = 1.0 x 10⁻¹⁴). Therefore, the pH (which is the negative logarithm of H⁺ concentration) and pOH (the negative logarithm of OH⁻ concentration) add up to 14, but H⁺ and OH⁻ themselves do not equal 14.
[H+] = Kw / [OH-] = 1.0*10-14 / 2.5*10-4 = 4.0*10-11 mol/L
-log[1 X 10^-4 M OH(-)] = 4 14 - 4 = 10 pH ----------------
The pH of a solution can be calculated using the formula pH = 14 - pOH. Given that the pOH is 3.31, we can subtract this value from 14 to find the pH. In this case, the pH of the solution would be approximately 10.69.
The pOH is 6,4.
To find the concentration of hydroxide ions ([OH-]) in a solution when the pH is 4.0, you can use the formula pH + pOH = 14. Since the pH is 4.0, the pOH would be 14 - 4 = 10. To convert pOH to [OH-] concentration, use the formula [OH-] = 10^(-pOH). Thus, [OH-] = 10^(-10) = 1 x 10^(-10) M.
The hydroxide ion concentration of a solution with pH 5.75 can be calculated using the formula [OH-] = 10^(-pOH). First, find the pOH by subtracting the pH from 14 (pOH = 14 - pH = 14 - 5.75 = 8.25). Then, calculate [OH-] = 10^(-8.25) ≈ 5.62 x 10^(-9) mol/L.
An acidic solution will have a pOH ranging from 7 to 14
pH and pOH are a measure of the concentration of the hydronium ions and hydroxyl ions respectively in the solution. pH = -log[H+] pOH = -log[OH-] and they are related: pH + pOH = 14
To find the pOH of a solution, you can use the formula pOH = -log[OH⁻]. Given that [OH⁻] = 1.41 × 10⁻¹³, calculate the pOH: pOH = -log(1.41 × 10⁻¹³) ≈ 12.85. Therefore, the pOH of the solution is approximately 12.85.
The pH of a 0.0110 M solution of Ba(OH)2 can be calculated by finding the hydroxide ion concentration, which is double the concentration of the Ba(OH)2 solution. Therefore, [OH-] = 2 * 0.0110 M = 0.0220 M. From this, you can calculate the pOH using the formula -log[OH-], and then convert pOH to pH using the relation pH + pOH = 14.
The sum of pH and pOH is always equal to 14 in a neutral solution at 25°C. This is because pH is a measure of the concentration of H+ ions in a solution while pOH is a measure of the concentration of OH- ions. In a neutral solution, the concentration of H+ ions is equal to the concentration of OH- ions, resulting in a sum of 14.
The pOH of a solution with an OH- concentration of 1.4 x 10^-13 M is 1.85. This is calculated by taking the negative logarithm of the OH- concentration, which is -log(1.4 x 10^-13) = 12.85, then subtracting it from 14 to find the pOH: 14 - 12.85 = 1.85.
To find the [H+] concentration in a solution with a pOH of 0.253, you first need to find the pOH of the solution which is 14 - pOH = 14 - 0.253 = 13.747. Then, you can use the relation [H+][OH-] = 1.0 x 10^-14 to calculate the [H+] concentration. [H+] = 10^-13.747 = 1.93 x 10^-14 M.
7.8