Consider a solution containing 0.47 M HOCl and 0.67 M NaOCl for the next two questions.
The Ka for HOCl = 3.5E-8.
What is the pH of this solution?
Now calculate the pH of this solution after 0.15 mol of NaOH is added to 1.00 L of the solution.
Assume that the volume does not change when the NaOH is added.
To calculate the pH of the solution containing 0.47 M HOCl and 0.67 M NaOCl, we need to consider the dissociation of HOCl as a weak acid.
1. pH Calculation:
HOCl dissociates in water according to the following equilibrium reaction:
HOCl ⇌ H+ + OCl-
The Ka value given for HOCl (3.5E-8) can be used to calculate the concentration of H+ ions in the solution.
We can set up an ICE table (Initial, Change, Equilibrium) to track the concentrations of the species involved in the equilibrium:
HOCl ⇌ H+ + OCl-
Initial (M): 0.47 0 0
Change (M): -x +x +x
Equilibrium: 0.47-x x x
Using the Ka expression for the reaction:
Ka = [H+][OCl-] / [HOCl]
Since the concentration of water remains constant, we can omit it from the expression.
Substituting the equilibrium concentrations into the Ka expression, we have:
Ka = (x)(x) / (0.47 - x) ≈ 3.5E-8
Since the initial concentration of HOCl (0.47 M) is much greater than the x value, we can assume that x is negligible compared to 0.47. Thus, we approximate 0.47 - x as 0.47 in the denominator.
Simplifying the equation, we have:
3.5E-8 = (x^2) / 0.47
Now, solve for x by rearranging the equation:
x^2 = (3.5E-8)(0.47)
x^2 = 1.645E-8
x ≈ 0.001282 M
The concentration of H+ ions is approximately 0.001282 M. Now, we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.001282)
pH ≈ 2.892
Therefore, the pH of this solution is approximately 2.892.
2. pH Calculation after adding 0.15 mol of NaOH to 1.00 L of the solution:
Since NaOH is a strong base, when it reacts with the weak acid HOCl, it will fully neutralize the H+ ions and produce the conjugate base OCl-. As a result, the concentration of H+ ions will decrease, leading to an increase in pH.
To calculate the new pH, we'll need to determine the new concentration of H+ ions.
Since 0.15 mol of NaOH is added to 1.00 L of the solution, NaOH is fully dissociated and will react with an equal number of H+ ions.
The number of moles of H+ ions neutralized by NaOH = 0.15 mol
The initial concentration of H+ ions is approximately 0.001282 M (as calculated earlier).
Since the volume does not change upon adding NaOH, the final concentration of H+ ions can be calculated using the formula:
Final concentration of H+ ions = (Initial concentration - moles of H+ neutralized) / Volume
Final concentration of H+ ions = (0.001282 - 0.15) / 1.00
Final concentration of H+ ions ≈ 0.001132 M
Finally, we can calculate the new pH using the formula:
pH = -log[H+]
pH = -log(0.001132)
pH ≈ 2.946
Therefore, after adding 0.15 mol of NaOH to 1.00 L of the solution, the new pH is approximately 2.946.