A buffer solution contains 0.283 M hypochlorous acid and 0.456 M potassium hypochlorite.
If 0.0464 moles of perchloric acid are added to 225 ml of this buffer, what is the pH of the resulting solution ?
(Assume that the volume change does not change upon adding perchloric acid)
pH =
To find the pH of the resulting solution after adding perchloric acid, we need to consider the reaction between perchloric acid (HClO4) and hypochlorous acid (HClO) in the buffer solution.
The balanced equation for the reaction is:
HClO4 + HClO → ClO4- + H2O
From the balanced equation, we can see that for every mole of perchloric acid (HClO4) that reacts, one mole of hypochlorous acid (HClO) is converted into hypochlorite ion (ClO-).
Given that 0.0464 moles of perchloric acid are added to the buffer solution, we need to calculate how many moles of hypochlorous acid (HClO) and hypochlorite ion (ClO-) remain in the solution after the reaction.
Initial moles of hypochlorous acid (HClO):
0.283 M × 0.225 L = 0.063675 moles
Initial moles of hypochlorite ion (ClO-):
0.456 M × 0.225 L = 0.1026 moles
Since the mole ratio between HClO4 and HClO is 1:1, 0.0464 moles of HClO4 react, and therefore, the final moles of HClO and ClO- are:
Final moles of hypochlorous acid (HClO):
0.063675 moles - 0.0464 moles = 0.017275 moles
Final moles of hypochlorite ion (ClO-):
0.1026 moles - 0.0464 moles = 0.0562 moles
Now that we have the concentrations of HClO and ClO- in the resulting solution as a result of the reaction, we can calculate the pH.
The Henderson-Hasselbalch equation relates pH, pKa, and the concentrations of a weak acid and its conjugate base:
pH = pKa + log([A-]/[HA])
In this case, HClO is the weak acid, and ClO- is its conjugate base. The pKa for the HClO is 7.53.
pH = 7.53 + log([ClO-]/[HClO])
= 7.53 + log(0.0562/0.017275)
≈ 11.479
Therefore, the pH of the resulting solution is approximately 11.479.
To determine the pH of the resulting solution, we need to consider the equilibrium that exists between hypochlorous acid (HOCl) and hypochlorite ion (OCl-) in the buffer solution.
The balanced chemical equation for the ionization of hypochlorous acid is as follows:
HOCl ⇌ H+ + OCl-
Given the initial concentrations of HOCl and OCl-, we can use the Henderson-Hasselbalch equation to calculate the pH.
The Henderson-Hasselbalch equation is:
pH = pKa + log([A-]/[HA])
Where pKa is the acid dissociation constant, [A-] represents the concentration of the conjugate base (OCl- in this case), and [HA] represents the concentration of the acid (HOCl in this case).
First, let's calculate the pKa value for hypochlorous acid (HOCl). The pKa value can be found in reference books or online databases. For hypochlorous acid, the pKa value is approximately 7.53.
Now, let's calculate the concentrations of [A-] and [HA] after the addition of perchloric acid.
After adding 0.0464 moles of perchloric acid to 225 ml of the buffer, the final volume becomes 225 mL. Since the volume change is not specified, we assume that it remains constant.
To calculate the concentration of OCl- ([A-]), we need to consider the reaction between perchloric acid (HClO4) and hypochlorite ion (OCl-) in the buffer:
HClO4 + OCl- → ClO4- + HOCl
Since 1 mole of perchloric acid (HClO4) reacts with 1 mole of hypochlorite ion (OCl-), the concentration of OCl- after the reaction is 0.0464 moles divided by the final volume (225 mL):
[A-] = 0.0464 moles / 225 mL
Similarly, the concentration of HOCl ([HA]) after the reaction is equal to its initial concentration minus the concentration of OCl- ([A-]).
[HA] = Initial concentration of HOCl - [A-]
[HA] = 0.283 M - [A-]
Now, substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 7.53 + log([A-]/[HA])
pH = 7.53 + log(0.0464 moles / 225 mL) / (0.283 M - [A-])
After calculating the value inside the logarithm and solving the equation, you will be able to find the pH of the resulting solution.
millimols HOCl = mL x M = 225 x 0.283 = estimated 64.
mmols KOCl = 225 x 0.456 = 103
add 46 mmols HClO4.
..........OCl^- + H^+ ==> HOCl + H2O
I.........64......0........103
add..............46.............
C........-46....-46.........+46
E.........18......0........149
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH. You will need to look up or calculate pKa for HOCl
Note that I have rounded on all of the numbers. You will need to recalculate and obtain more accurate numbers.