Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.240 M HClO(aq) with 0.240 M KOH(aq). The ionization constant for HClO can be found here.
Divide the titration curve into 4 parts.
a. beginning
b. between beginning and eq pt.
c. eq pt
d. after eq pt.
First, determine mL to eq pt.
mLacid x M aci = mL base x M base and that determination tells you how to divide into the four part.
a.......... HClO ==> H^+ + ClO^-
I..........0.240M.....0.....0
C............-x.......x......x
E.........0.240-x.....x......x
Substitute the E line into the Ka expression for HClO and solve for x = (H^+), then convert to pH.
b. mmols to start = mL x M = 50*0.240 = about 12.
mmols KOH added. You don't give a value but lets say 25 mL. So 25 x 0.240 = about 6. Therefore, you have 6 acid left and you formed 6 of the salt from
.......HClO + KOH ==> KClO + H2O
I.......12.....0........0......0
add............6.............
C.......-6.....-6........6......6
E........6.......0.......6.......6
Use the Henderson-Hasselbalch equation to solve for the pH.
pH = pKa for HClO + log (6/6)
?
c. The pH is determined by the hydrolysis of the salt.
........ClO^- + HOH ==>HClO + OH^-
I......0.12............0......0
C........-x............x......x
E........-0.12-x........x.....x
Kb for ClO^- = (Kw/Ka for HClO) = (x*x)/(0.12-x).
Substitute and solve for x = OH^- and convert to pH.
NOTE: The concn of the salt at the eq.pt is just 1/2 of 0.240. Since they are the same molarity you have added 50 mL KOH so the M salt is just 1/2 of what acid or base started.
d. After the eq. pt is just excess KOH.
To calculate the pH at different points during the titration of a strong acid (HClO) with a strong base (KOH), we need to understand the reaction between HClO and KOH. The balanced equation for this reaction is:
HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)
At the beginning of the titration, before any KOH is added, we have only the HClO solution. It is assumed that HClO dissociates completely, so we can consider it as a strong acid. In water, HClO completely ionizes to release H+ ions, resulting in the formation of ClO- ions. The ionization reaction is:
HClO(aq) → H+(aq) + ClO-(aq)
Since we are given the ionization constant (Ka) for HClO, we can use it to calculate the H+ concentration, and thus the pH.
The ionization constant (Ka) for HClO is a measure of the strength of the acid. It can be found in the given reference or in a chemistry textbook. For HClO, the value of Ka is 3.0 × 10^-8.
Using the concentration of HClO (0.240 M) and the ionization constant (Ka), we can set up an ICE table to find the H+ concentration.
Initial:
[HClO] = 0.240 M
[H+] = 0
Change:
[HClO] = -x
[H+] = +x
Equilibrium:
[HClO] = 0.240 - x
[H+] = x
Since HClO completely ionizes, at equilibrium, the concentration of HClO is reduced by the same amount as the concentration of H+ ions formed. Therefore, [HClO] can be approximated as 0.240 M, and [H+] can be approximated as x.
Using the equation for Ka, we have:
Ka = [H+][ClO-] / [HClO]
Rewriting in terms of x:
3.0 × 10^-8 = x * (0.240 - x) / 0.240
Simplifying:
3.0 × 10^-8 = x * (1 - x / 0.240)
Solving the quadratic equation:
x^2 - 0.240x + 0.240 * 3.0 × 10^-8 = 0
Using the quadratic formula, we get two possible values for x: x ≈ 5.5 × 10^-5 or x ≈ 0.240.
Since the value of x must be smaller than the initial concentration of HClO (0.240 M), we select the value x ≈ 5.5 × 10^-5.
So, the H+ concentration at the beginning of the titration is approximately 5.5 × 10^-5 M. To find the pH, we use the equation pH = -log[H+].
pH = -log(5.5 × 10^-5) ≈ 4.26
Therefore, the pH at the beginning of the titration is approximately 4.26.
To calculate the pH at different points during the titration of HClO with KOH, we need to determine the number of moles of each species present at each point and then use these quantities to calculate the concentration of H+ ions.
First, let's write the balanced equation for the reaction between HClO and KOH:
HClO(aq) + KOH(aq) → KClO(aq) + H2O(l)
At the start of the titration (before any KOH is added), we have 50.0 mL * 0.240 M = 12.0 mmol HClO. Since HClO is a weak acid, it partially ionizes in water:
HClO(aq) + H2O(l) ⇌ H3O+(aq) + ClO-(aq)
The ionization constant for HClO (Ka) is given in the problem statement.
Now, let's look at the different points during the titration:
1. Before any KOH is added (before the equivalence point):
At this point, there are only the initial 12.0 mmol of HClO present. Therefore, the concentration of HClO is 12.0 mmol / 50.0 mL = 0.240 M.
To calculate the pH, we need to determine the concentration of H3O+ using the equation for the ionization of HClO.
Ka = [H3O+][ClO-] / [HClO]
Since the initial concentration of ClO- is negligible compared to the initial concentration of HClO, we can assume that [ClO-] = 0. Therefore:
Ka = [H3O+][0] / [HClO]
[H3O+] = (Ka * [HClO])^0.5
[H3O+] = (Ka * 0.240)^0.5
[H3O+] ≈ (3.5 * 10^-8 * 0.240)^0.5
[H3O+] ≈ (8.4 * 10^-9)^0.5
[H3O+] ≈ 9.2 * 10^-5 M
pH = -log([H3O+])
pH = -log(9.2 * 10^-5)
pH ≈ 4.04
Therefore, before any KOH is added (before the equivalence point), the pH is approximately 4.04.
2. At the equivalence point:
The equivalence point occurs when the moles of KOH added is equal to the moles of HClO present initially. Since we have 12.0 mmol of HClO, it will take 12.0 mmol of KOH to reach this point.
At this point, all of the HClO has reacted, and we only have KClO and water. The resulting solution will be the neutral salt KClO, which does not affect the pH. Therefore, the pH at the equivalence point is approximately 7 (neutral).
Please note that this calculation assumes the ionization constant (Ka) provided is valid for the given temperature and concentration range.