- log(1 X 10^-5 M)
= 5
14 - 5
= 9 pH
----------
[OH-] = 1x10^-4.22 or more conventionally, [OH-] = 6.03x10^-5 M
The concentration of OH- for a solution with H3O+ concentration of 1x10^-5 M can be found by using the ion product constant of water (Kw = 1.0x10^-14) to calculate the OH- concentration. Since H3O+ and OH- are related by Kw = [H3O+][OH-], you can solve for [OH-] by rearranging the equation. This will give you a value of 1.0x10^-9 M for the OH- concentration.
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 pH level when [OH-] equals 2.3 x 10^-3 M can be calculated using the equation: pOH = -log[OH-]. Given [OH-] = 2.3 x 10^-3 M, pOH = -log(2.3 x 10^-3) ≈ 2.64. Since pH + pOH = 14, pH = 14 - pOH = 14 - 2.64 ≈ 11.36.
To calculate the concentration of hydroxide ions (OH-) from a given pH value, you can use the formula: [OH-] = 10^(-pH). For a pH of 1.12, the concentration of hydroxide ions would be [OH-] = 10^(-1.12) = 0.079 moles per liter.
- log(1 X 10 -12 M) = 12 pH -----------------------very little room for H3O
[OH-] = 1x10^-4.22 or more conventionally, [OH-] = 6.03x10^-5 M
[OH-] = 1x10^-3 M[H+][OH-] = 1x10^-14[H+] = 1x10^-14/1x10^-3 = 1x10^-11pH = -log 1x10^-11 = 11Done another way:pOH = -log [OH-] = -log 1x10^-3 = 3pH + pOH = 14pH = 14 - 3 = 11
[OH-] = 1x10^-4.22 or more conventionally, [OH-] = 6.03x10^-5 M
In pure water, the hydrogen ion (H+) concentration is 1x10^-7 M and the hydroxide ion (OH-) concentration is also 1x10^-7 M. The pH of pure water is 7 (neutral) and the pOH is also 7.
- log(1 X 10 -3 M H +) = 3 pH =====
if 0.000001 is the hydrogen ion concentration the pH is 6
The pH of a solution can be calculated using the formula pH = 14 - (-log[OH-]). Using the given concentration of 2.3 x 10^-5 M for OH-, the pH would be approximately 9.64.
The concentration of OH- for a solution with H3O+ concentration of 1x10^-5 M can be found by using the ion product constant of water (Kw = 1.0x10^-14) to calculate the OH- concentration. Since H3O+ and OH- are related by Kw = [H3O+][OH-], you can solve for [OH-] by rearranging the equation. This will give you a value of 1.0x10^-9 M for the OH- concentration.
The pH of a 10^-5 M KOH solution would be around 9. For a given concentration of a strong base like KOH, the pH can be calculated using the formula pH = 14 - pOH. Given that pOH = -log[OH-] and [OH-] = 10^-5 M in this case, pOH = 5. Therefore, pH = 14 - 5 = 9.
A solution with a hydroxide-ion concentration of 1x10-4 M is considered basic because it has a higher concentration of hydroxide ions (OH-) relative to hydronium ions (H+). This concentration indicates a low pH value and a high alkalinity.
A micron (µm), presumably referring to a Micrometre is one millionth of a meter (1x10-6 m ). Thus, making 10 microns (1x10-5 m). A micron (µm), presumably referring to a Micrometre is one millionth of a meter (1x10-6 m ). Thus, making 10 microns (1x10-5 m).