To find the concentrations of H3O+ and OH- at body temperature (37°C), we can use the equation for Kw:
Kw = [H3O+][OH-]
Given that Kw = 1.00 x 10^-13.60, we can substitute this value into the equation:
1.00 x 10^-13.60 = [H3O+][OH-]
Since water is neutral, the concentration of H3O+ is equal to the concentration of OH-. Let's denote this concentration as x:
x * x = 1.00 x 10^-13.60
Taking the square root of both sides, we get:
x = √(1.00 x 10^-13.60)
Calculating this, we find:
x ≈ 1.00 x 10^-6.80
Therefore, the concentration of H3O+ and OH- at body temperature (37°C) is approximately 1.00 x 10^-6.80 M.
The ionization of water is kW=1.00x10^-13.60 at body temperature 37°C .what are the H3O^+ and OH^- concentration at the temperature?
5 answers
Calculate the pH of 0.025M solution ammonia, Kb =1.75x10^-5
Ammonia (NH3) is a weak base, and we are given the base dissociation constant (Kb) as 1.75x10^-5.
To calculate the pH of the 0.025M ammonia solution, we need to determine the concentration of hydroxide ions (OH-) in the solution.
First, let's calculate the concentration of hydroxide ions (OH-) using Kb:
Kb = [NH4+][OH-] / [NH3]
Since ammonia is a weak base, we assume that the concentration of NH3 will remain essentially unchanged after it dissociates. Therefore, we can assume the concentration of NH3 is equal to the initial concentration, which is 0.025M.
1.75x10^-5 = x^2 / 0.025
Solving for x, we find:
x ≈ √(1.75x10^-5 * 0.025)
x ≈ √(4.375x10^-7)
x ≈ 6.609x10^-4
This is the concentration of hydroxide ions (OH-) in the solution.
To calculate the pOH, we can use the formula:
pOH = -log10[OH-]
pOH = -log10(6.609x10^-4)
pOH ≈ -(-3.18)
pOH ≈ 3.18
Finally, to find the pH, we can use the formula:
pH = 14 - pOH
pH ≈ 14 - 3.18
pH ≈ 10.82
Therefore, the pH of the 0.025M ammonia solution is approximately 10.82.
To calculate the pH of the 0.025M ammonia solution, we need to determine the concentration of hydroxide ions (OH-) in the solution.
First, let's calculate the concentration of hydroxide ions (OH-) using Kb:
Kb = [NH4+][OH-] / [NH3]
Since ammonia is a weak base, we assume that the concentration of NH3 will remain essentially unchanged after it dissociates. Therefore, we can assume the concentration of NH3 is equal to the initial concentration, which is 0.025M.
1.75x10^-5 = x^2 / 0.025
Solving for x, we find:
x ≈ √(1.75x10^-5 * 0.025)
x ≈ √(4.375x10^-7)
x ≈ 6.609x10^-4
This is the concentration of hydroxide ions (OH-) in the solution.
To calculate the pOH, we can use the formula:
pOH = -log10[OH-]
pOH = -log10(6.609x10^-4)
pOH ≈ -(-3.18)
pOH ≈ 3.18
Finally, to find the pH, we can use the formula:
pH = 14 - pOH
pH ≈ 14 - 3.18
pH ≈ 10.82
Therefore, the pH of the 0.025M ammonia solution is approximately 10.82.
Calculate the pH value of 0.001mol solution of NaoH at 25°C
To calculate the pH value of a sodium hydroxide (NaOH) solution, we need to determine the concentration of hydroxide ions (OH-) in the solution.
For a 0.001 mol solution of NaOH, we know that it fully dissociates into Na+ and OH- ions. Therefore, the concentration of OH- ions is equal to the concentration of NaOH, which is 0.001 mol/L.
Next, we can use the formula for calculating pOH:
pOH = -log10[OH-]
pOH = -log10(0.001)
pOH = -(-3)
pOH = 3
Finally, we can calculate the pH value using the formula:
pH = 14 - pOH
pH = 14 - 3
pH = 11
Therefore, the pH value of a 0.001 mol sodium hydroxide solution at 25°C is 11.
For a 0.001 mol solution of NaOH, we know that it fully dissociates into Na+ and OH- ions. Therefore, the concentration of OH- ions is equal to the concentration of NaOH, which is 0.001 mol/L.
Next, we can use the formula for calculating pOH:
pOH = -log10[OH-]
pOH = -log10(0.001)
pOH = -(-3)
pOH = 3
Finally, we can calculate the pH value using the formula:
pH = 14 - pOH
pH = 14 - 3
pH = 11
Therefore, the pH value of a 0.001 mol sodium hydroxide solution at 25°C is 11.