Calculate the equilibrium concentration of all ionic aqueous species in a solution containing 1.30 M HCl and 4.0 M C6H5OH (phenol, Ka=1.6 x 10^-10). What is the pH of the solution?
How about showing how much already know about the problem? And when you get as far as you can go, please explain what you don't understand about the next step.
I was just wondering whether HCl and C6H5OH on the reactant side together, or am i supposed to separate the two into two different reaction equations, and then do ICE from there? Because I'm not sure how to create a acid-base reaction with HCl and C6H5OH together on one side... I was thinking separating at first .. HCl => H+ + Cl-, but that seems wrong also
Separate them. There is no reaction between them. Use Ka from phenol to include (H^+) from HCl and calculate total H+.
To calculate the equilibrium concentration of ionic aqueous species in a solution, we need to consider the dissociation of the acidic substance (HCl) and the weak acid (phenol, C6H5OH).
1. For HCl:
HCl is a strong acid, which means it dissociates completely in water. Therefore, the concentration of H+ ions in the solution will be equal to the initial concentration of HCl, which is 1.30 M.
2. For phenol (C6H5OH):
Phenol is a weak acid with the dissociation reaction:
C6H5OH ⇌ C6H5O- + H+
The dissociation of the weak acid can be expressed with the ionization constant (Ka), which is given as 1.6 x 10^-10.
Let's assume x M is the equilibrium concentration (also the concentration of H+ ions) formed from the dissociation of phenol. Then, the equilibrium concentration of the C6H5O- ion will also be x M.
Using the expression for the ionization constant, we can write:
Ka = [C6H5O-][H+] / [C6H5OH]
1.6 x 10^-10 = (x)(x) / (4.0)
1.6 x 10^-10 = (x^2) / (4.0)
Solving this equation, we find that x is approximately 2.53 x 10^-6 M. Hence, the equilibrium concentration of both H+ and C6H5O- ions is approximately 2.53 x 10^-6 M.
Now, to calculate the pH of the solution, we can use the equation pOH = -log10 [OH-]. Since the concentration of OH- ions is negligible compared to H+ and C6H5O- ions, we can assume the concentration of OH- ions to be close to zero.
Therefore, the concentration of H+ ions is equal to the concentration of HCl (1.30 M) + the concentration of H+ ions formed from the dissociation of phenol (2.53 x 10^-6 M). So, the total concentration of H+ ions is approximately 1.30 M + 2.53 x 10^-6 M = 1.30000253 M.
Finally, to calculate the pH, we can use the equation pH = -log10 [H+]. Substituting the concentration of H+ ions, we find that the pH of the solution is approximately -log10 (1.30000253) = 0.884.
Therefore, the pH of the solution is approximately 0.884.