Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.150 M HClO(aq) with 0.150 M KOH(aq). The ionization constant for HClO is 4.0x10^-8.
pH before the addition of any KOH?
pH after the addition of 25 mL of KOH?
Please show the steps to solving this problem.
To calculate the pH for each case in the titration, we need to understand the process and use the appropriate formulas. Let's go step by step to solve each part of the problem.
1. pH before the addition of any KOH:
Before the addition of any KOH, we only have HClO present. We are given that the initial concentration of HClO is 0.150 M. Since HClO is a weak acid, we can use the ionization constant (Ka) to calculate the concentration of H+ ions.
The ionization constant (Ka) for HClO is 4.0x10^-8. It represents the equilibrium equation:
HClO ⇌ H+ + ClO-
The equilibrium expression can be written as:
Ka = [H+][ClO-] / [HClO]
Since [H+][ClO-] represents the concentration of H+ ions, and assuming the initial concentration of H+ is x, we can rewrite the equilibrium expression as:
Ka = x^2 / (0.150 - x)
Since HClO is a weak acid, we can assume that x is small compared to the initial concentration, so we can neglect it in the denominator:
Ka = x^2 / 0.150
Rearranging the equation to solve for x, we get:
x^2 = Ka * 0.150
x = √(Ka * 0.150)
Substituting the value of Ka, we find:
x = √(4.0x10^-8 * 0.150)
x ≈ 1.5493x10^-5 M
Now that we have the concentration of H+ ions, we can calculate the pH using the formula:
pH = -log[H+]
Substituting the value of [H+], we get:
pH = -log(1.5493x10^-5)
pH ≈ 4.81
Therefore, before the addition of any KOH, the pH is approximately 4.81.
2. pH after the addition of 25 mL of KOH:
Since we know the initial volume and concentration of KOH added, we can calculate the moles of KOH added using the equation:
moles KOH = concentration KOH * volume KOH
moles KOH = 0.150 M * 0.025 L
moles KOH = 0.00375 mol
Since 1 mole of KOH reacts with 1 mole of HClO in a 1:1 ratio, the moles of HClO will be equal to the moles of KOH added.
Now we can calculate the new volume of the solution after the addition of KOH. Since we started with 50.0 mL of HClO and added 25 mL of KOH, the total volume will be 75 mL (0.075 L).
To find the new concentration of HClO, divide the moles of HClO by the new volume:
concentration HClO = moles HClO / volume solution
concentration HClO = 0.00375 mol / 0.075 L
concentration HClO = 0.05 M
Now, we can repeat the same process as before to calculate the new concentration of H+ ions and the pH using the new concentration of HClO.
Using the same equilibrium expression as before:
Ka = [H+][ClO-] / [HClO]
We have the new concentration of HClO and calculate the new concentration of H+ ions (x) as before.
Substituting the new concentration of HClO and solving for x, we get:
x = √(Ka * 0.05)
Substituting the value of Ka, we find:
x = √(4.0x10^-8 * 0.05)
x ≈ 1.4142x10^-5 M
Now we can calculate the pH using the same formula as before:
pH = -log[H+]
Substituting the value of [H+], we get:
pH = -log(1.4142x10^-5)
pH ≈ 4.85
Therefore, after the addition of 25 mL of KOH, the pH is approximately 4.85.
To calculate the pH for each case in the titration, we need to determine the number of moles of HClO and KOH, as well as the resulting concentrations of HClO and ClO- at each step.
1. pH before the addition of any KOH:
Before the titration begins, only HClO is present. The pH is calculated using the concentration of HClO and the ionization constant (Ka) of HClO.
Step 1: Calculate the number of moles of HClO:
moles HClO = volume (L) x concentration (M)
moles HClO = 0.050 L x 0.150 M
moles HClO = 0.0075 mol
Step 2: Calculate the concentration of HClO:
[HClO] = moles of HClO / volume of solution (L)
[HClO] = 0.0075 mol / 0.050 L
[HClO] = 0.15 M
Step 3: Calculate the pH using the ionization constant (Ka):
Ka = [H+][ClO-] / [HClO]
[HClO] = 0.15 M
[ClO-] = 0 M (since no KOH has been added)
[H+] = ? (what we want to find)
Ka = (x)(0) / 0.15 (since [ClO-]=0)
4.0x10^-8 = 0 / 0.15
4.0x10^-8 = x
pH = -log [H+]
pH = -log (4.0x10^-8)
pH ≈ 7.40
Therefore, the pH before the addition of any KOH is approximately 7.40.
2. pH after the addition of 25 mL of KOH:
At this point, half of the HClO has reacted with KOH to form a buffer solution. We need to determine the concentrations of HClO and ClO- after this reaction.
Step 1: Calculate the number of moles of HClO remaining:
moles HClO remaining = initial moles HClO - moles HClO reacted
moles HClO remaining = 0.0075 mol - (0.150 M x 0.025 L)
moles HClO remaining = 0.00375 mol
Step 2: Calculate the remaining concentration of HClO:
[HClO] = moles HClO remaining / volume of solution (L)
[HClO] = 0.00375 mol / 0.075 L (total volume after addition of KOH)
[HClO] = 0.05 M
Step 3: Calculate the concentration of ClO- (result of HClO reacting with KOH):
[ClO-] = moles of ClO- formed / volume of solution (L)
[ClO-] = moles KOH added / volume of solution (L)
[ClO-] = (0.150 M x 0.025 L) / 0.075 L
[ClO-] = 0.05 M
Step 4: Calculate the pH using the ionization constant (Ka):
Ka = [H+][ClO-] / [HClO]
[HClO] = 0.05 M
[ClO-] = 0.05 M
[H+] = ? (what we want to find)
Ka = (x)(0.05) / 0.05
4.0x10^-8 = x
pH = -log [H+]
pH = -log (4.0x10^-8)
pH ≈ 7.40
Therefore, the pH after the addition of 25 mL of KOH is approximately 7.40.
HClO + KOH ==> KClO + H2O
At the beginning you have pure HClO.
.......HClO ==> H^+ + ClO^-
I.....0.150......0......0
C........-x......x......x
E...0.150-x......x......x
Substitute the E line into the Ka expression and solve for x = H^+ and convert to pH.
After reaction:
HClO + KOH ==> KClO + H2O
mols HClO initially = M x L = ?
mols KOH added = M x L = ?
How much of the HClO is used? how much is left?
How much of the KOH has been used? How much is left.
You have prepared a buffer solution at this point and the pH can be determined by using te Henderson-Hasselbalch equation. Substitute and solve for pH. Post your work if you get stuck.