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?
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.
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.
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.
To determine the H3O+ and OH- concentration in water at body temperature (37°C) using the given ionization constant (Kw=1.00x10^-13.60), we can make use of the equation for Kw:
Kw = [H3O+][OH-]
At equilibrium, the concentration of H3O+ is equal to the concentration of OH-, denoted by x for both.
Substituting the values into the equation:
1.00x10^-13.60 = x * x
To solve this equation, we can take the square root of both sides:
√(1.00x10^-13.60) = √(x * x)
1.00x10^-13.60 = x
Now we have the concentration, x, for both H3O+ and OH- ions at 37°C in water, which is approximately 1.00x10^-13.60.
To find the H3O+ and OH- concentration at a given temperature, we can use the ionization constant of water (Kw) and the equation for the ionization of water:
H2O(l) ⇌ H+(aq) + OH-(aq)
At body temperature (37°C), we can assume that the water is in a state of dynamic equilibrium, where the concentrations of H3O+ and OH- are constant.
The ionization constant of water (Kw) is defined as the product of the concentrations of H3O+ and OH- ions in water. So, mathematically:
Kw = [H3O+][OH-]
At 37°C, the value of Kw is given as 1.00x10^-13.60.
To find the concentration of H3O+ and OH-, we need to solve for them. However, note that the concentrations of H3O+ and OH- in pure water are equal, as they are formed in a 1:1 ratio. Let's assume that the concentration of H3O+ and OH- in pure water is x.
Now, we can substitute the value of Kw and the assumption of equal concentrations into the equation:
Kw = [H3O+][OH-]
1.00x10^-13.60 = x * x
1.00x10^-13.60 = x^2
To solve for x, we take the square root of both sides:
√(1.00x10^-13.60) = √(x^2)
0.316 = x
So, the concentration of both H3O+ and OH- in pure water at 37°C is 0.316 (assuming the units are Molarity or mol/L).
Therefore, the H3O+ concentration and OH- concentration at body temperature are both 0.316 M.