The pH of pure water at 50 degrees celsius is 6.63. What is the value of Kw at 50 degrees celsius?
To find the value of Kw (ionic product of water) at 50 degrees Celsius, we can use the equation:
Kw = [H+][OH-]
In pure water, the concentration of H+ ions is equal to the concentration of OH- ions since pure water is neutral. Therefore:
[H+] = [OH-]
Let x denote the concentration of H+ and OH- ions in pure water at 50 degrees Celsius.
From the given pH of pure water (6.63), we can find the H+ concentration using the equation:
[H+] = 10^(-pH)
[H+] = 10^(-6.63)
Now, since [H+] = [OH-], the value of Kw can be calculated as follows:
Kw = [H+][OH-] = x * x = x^2
So, to find the value of Kw, we need to find the concentration of H+ ions first.
[H+] = 10^(-6.63) ≈ 2.19 × 10^(-7)
Since [H+] = [OH-], the concentration of OH- ions is also 2.19 × 10^(-7)
Now we can calculate Kw:
Kw = (2.19 × 10^(-7)) * (2.19 × 10^(-7))
Kw ≈ 4.80 × 10^(-14)
Therefore, the value of Kw at 50 degrees Celsius is approximately 4.80 × 10^(-14).
To find the value of Kw (the ion product constant for water) at a specific temperature, we need to use the equilibrium constant expression for water dissociation:
Kw = [H+][OH-]
At 25 degrees Celsius, the standard temperature, the value of Kw is known to be 1.0 x 10^-14. However, since we are interested in the value of Kw at 50 degrees Celsius, we need to consider the temperature dependence of Kw.
The temperature dependence of Kw can be described by the equation:
Kw2 = Kw1 × (T2 / T1)^(Δn)
Where:
- Kw2 is the value of Kw at the desired temperature (50 degrees Celsius)
- Kw1 is the value of Kw at the standard temperature (25 degrees Celsius)
- T2 is the desired temperature in Kelvin (323 K)
- T1 is the standard temperature in Kelvin (298 K)
- Δn represents the change in the number of moles of gaseous products minus the number of moles of gaseous reactants. For water, the change is zero since there are equal moles of products (H+ and OH-) and reactants (H2O).
Let's apply this equation to calculate the value of Kw at 50 degrees Celsius:
T2 = 50 degrees Celsius + 273.15 K = 323.15 K
T1 = 25 degrees Celsius + 273.15 K = 298.15 K
Δn = 0 (since the number of moles of H+ and OH- is the same as the number of moles of H2O)
Kw2 = Kw1 × (T2 / T1)^(Δn)
= 1.0 x 10^-14 × (323.15 K / 298.15 K)^(0)
= 1.0 x 10^-14
Therefore, the value of Kw at 50 degrees Celsius is still 1.0 x 10^-14, the same as at 25 degrees Celsius.
pH = -log(H^+)
-6.63 = log(H^+)
(H^+) = 2.34E-7
For pure water, H2O ==> H^+ + OH^-
Kw = (H^+)(OH^-)
Kw = (2.34E-7)(2,34E-7)