at 25 celcius, how many dissociated OH ions are there in 8.0x10^-1 L of aqueous solution with pH 4.73?
H2O ==> H^+ + OH^-
Kw = (H^+)(OH^-)
pH = 4.73 = -log(H^+) and I get approx 2E-5 but that's just a close estimate for (H^+). That makes (OH^-) = estimated 5E-10M
# mols OH^- = M x L = ? and each mols has 6.02E23 ions.
To calculate the number of dissociated OH- ions in the given aqueous solution, we need to first determine the concentration of OH- ions.
The pH of a solution is related to the concentration of H+ ions using the equation: pH = -log[H+]. Rearranging the equation gives us [H+] = 10^(-pH).
Since the solution is aqueous, we can assume it has a neutral pH of 7. In this case, the concentration of H+ ions would be [H+] = 10^(-pH) = 10^(-7) = 1x10^(-7) M.
Now, we know that water undergoes auto-ionization to produce equal concentrations of H+ and OH- ions in pure water. Therefore, the concentration of OH- ions can be considered to be the same as the H+ concentration at 25 degrees Celsius.
To calculate the number of moles of OH- ions in the given volume of the solution, we multiply the concentration by the volume:
Moles of OH- ions = concentration of OH- ions (M) × volume of the solution (L).
Moles of OH- = (1x10^(-7) M) × (8.0x10^(-1) L) = 8.0x10^(-8) mol.
However, this calculation gives us the total moles of OH- ions, including both dissociated and non-dissociated OH- ions. In an aqueous solution, only a small fraction of OH- ions dissociate, while the remaining form water molecules through recombination. The degree of dissociation can be determined using the pOH of the solution.
The pOH is the negative logarithm of the concentration of OH- ions: pOH = -log[OH-].
Given that the pH of the solution is 4.73, the pOH can be obtained as follows: pOH = 14 - pH = 14 - 4.73 = 9.27.
The concentration of OH- ions can be calculated using the equation: [OH-] = 10^(-pOH).
[OH-] = 10^(-9.27) = 1.93x10^(-10) M.
Now, we can determine the proportion of dissociated OH- ions in the solution. To do this, we need to compare the concentration of OH- ions with the water auto-ionization constant, Kw, which is equal to 1.0x10^(-14) at 25 degrees Celsius.
The concentration of dissociated OH- ions can be calculated as follows:
Concentration of dissociated OH- ions = [OH-] - (Kw / [OH-]).
Concentration of dissociated OH- = 1.93x10^(-10) M - (1.0x10^(-14) M) / (1.93x10^(-10) M) = 1.93x10^(-10) M.
Finally, we multiply the concentration of dissociated OH- ions by the volume of the solution to get the number of dissociated OH- ions:
Number of dissociated OH- ions = concentration of dissociated OH- ions (M) x volume of the solution (L).
Number of dissociated OH- ions = (1.93x10^(-10) M) × (8.0x10^(-1) L) = 1.54x10^(-10) mol.
So, there are approximately 1.54x10^(-10) moles of dissociated OH- ions in 8.0x10^(-1) L of the given aqueous solution.