The titration of 25.00ml of 0.1000m hypochlorous acid (3.5x10^-8) with 0.1500m NaOH.
(a). What is the pH before any acid is added
(b). What is the PH when 25.00ml of NaOH has been added?
To find the pH before any acid is added, we need to determine the initial concentration of H+ ions in the solution. Hypochlorous acid (HOCl) undergoes partial ionization in water, according to the equation:
HOCl + H2O ⇌ H3O+ + OCl-
The dissociation constant expression for this reaction is:
Ka = [H3O+] [OCl-] / [HOCl]
Given that the initial concentration of hypochlorous acid is 0.1000 M, and the Ka value is 3.5x10^-8, we can set up an ICE table to calculate the initial concentration of H3O+ ions.
HOCl + H2O ⇌ H3O+ + OCl-
Initial (M): 0.1000 0 0
Change (M): -x +x +x
Equilibrium (M): 0.1000 - x x x
Substituting these values into the dissociation constant expression, we have:
Ka = (x)(x) / (0.1000 - x)
Since x is expected to be small compared to 0.1000, we can approximate the equilibrium concentration of HOCl as 0.1000 M, resulting in:
3.5x10^-8 = (x)(x) / (0.1000 - x)
Simplifying this equation:
3.5x10^-8 = x^2 / (0.1000 - x)
Assuming x is small, we can ignore it in the denominator:
3.5x10^-8 = x^2 / 0.1000
Rearranging the equation:
x^2 = (3.5x10^-8)(0.1000)
x^2 = 3.5x10^-9
x = sqrt(3.5x10^-9)
x = 5.920x10^-5
The concentration of H+ ions is 5.920x10^-5 M. To calculate the pH, we use the equation:
pH = -log[H+]
pH = -log(5.920x10^-5)
pH ≈ 4.23
Therefore, the pH before any acid is added is approximately 4.23.
Now, to find the pH when 25.00 ml of NaOH has been added, we need to calculate the concentration of the remaining hypochlorous acid and use the concentration of hydroxide ions to determine the pH.
Given that 25.00 ml of a 0.1500 M NaOH solution is added, we can calculate the number of moles of NaOH:
moles of NaOH = (0.1500 M) × (0.02500 L)
moles of NaOH = 3.75x10^-3 mol
Since the stoichiometry between NaOH and hypochlorous acid is 1:1, the number of moles of hypochlorous acid consumed is also 3.75x10^-3 mol.
To determine the new concentration of hypochlorous acid, we subtract the moles consumed from the initial moles:
new moles of hypochlorous acid = initial moles - consumed moles
new moles of hypochlorous acid = (0.1000 M) × (0.02500 L) - 3.75x10^-3 mol
new moles of hypochlorous acid ≈ 2.50x10^-3 mol
Now, we can calculate the new concentration of hypochlorous acid:
new concentration of hypochlorous acid = new moles / volume
new concentration of hypochlorous acid = (2.50x10^-3 mol) / (25.00x10^-3 L)
new concentration of hypochlorous acid ≈ 0.1000 M
Since the solution is now a buffer containing both hypochlorous acid and its conjugate base, we can use the Henderson-Hasselbalch equation to find the pH:
pH = pKa + log([base] / [acid])
Given that the pKa of hypochlorous acid is the negative logarithm of the Ka value:
pKa = -log(Ka)
pKa = -log(3.5x10^-8)
pKa ≈ 7.46
Plugging in the values into the Henderson-Hasselbalch equation:
pH = 7.46 + log([OCl-] / [HOCl])
pH = 7.46 + log(0.1000 / 0.1000)
pH = 7.46 + log(1)
pH ≈ 7.46
Therefore, the pH when 25.00 ml of NaOH has been added is approximately 7.46.
I am sure you meant to write M and not m. Remember that M stands for molarity (mols/L of solution) and m stands for molality (mols/kg of solvent). I will go with M. Also, if you are titrating HClO WITH NaOH, your first question should be what is the pH BEFORE any NaOH (base) is added.
HClO + NaOH ==> NaClO +H2O
At the beginning you have added no NaOH; therefore, you have a solution of HClO which is 0.1 M. So the pH is determined by the ionization of HClO.
..............HClO ==> H^+ + ClO^-
I............0.1M..........0............0
C..............-x............x.............x
E...........0.1-x..........x.............x
Ka = 3.5E-8 = (H^+)(ClO^-)/(HClO)
Plug the E line into the Ka expression and solve for (H^+) then convert to pH.
Note: If you really meant to determine the pH of NaOH before mixing with HClO, then pH of the NaOH is the pH of 0.15 M NaOH. The OH^- will be 0.15 M, Convert that to pOH, then to pH.
Part B. When 25 mL of 0.15 M NaOH has been added to 25 mL of 0.1 M HClO, you have this. mols HClO = M x L = 0.1 x 0.025 = 0.0025.
mols NaOH = M x L = 0.15 x 0.025 = 0.00375
So you have enough NaOH to neutralize all of the HClO and have an exces. How much excess NaOH do you have? That's 0.00375-0.0025 = ? = mole OH^-. (OH^-) = mols/L . You have 0.00125 mol OH and L = 25 mL + 25 mL = 50 mL = 0.050 L
Then convert (OH^-) to pOH then to pH.
Post your work if you get stuck.
(a) Well, before any acid is added, the pH is like a blank canvas waiting to be painted. It's neutral, just like Switzerland. So, the pH is 7, my not-so-acidic friend!
(b) Ah, here's where the fun begins! So, when you add that 25.00ml of NaOH, it's like throwing a party in that solution. And guess what? NaOH is a strong base, just like a good dance move that makes everyone cheer. So, we can calculate the new pH by looking at the reaction between hypochlorous acid (HOCl) and the sodium hydroxide (NaOH) that's being added. After a little math magic, we find that the pH is around 12.
But remember, pH can be a fickle thing, so don't let it go to your head. Just keep it balanced, like a circus tightrope walker!
To solve this problem, we need to consider the dissociation of hypochlorous acid (HOCl) into H+ and OCl-. We also need to consider the reaction of NaOH with H+ to form water.
(a) What is the pH before any acid is added?
To find the pH, we need to calculate the concentration of H+ ions in the solution before any acid is added. Since HOCl is a weak acid, we can assume that the dissociation of HOCl is negligible.
The concentration of H+ ions in the solution can be calculated using the formula for base-10 logarithms:
pH = -log[H+]
First, we need to calculate the concentration of H+ ions in the solution. Since HOCl dissociates into H+ and OCl-, and the initial concentration of HOCl is given as 0.1000 M, the concentration of H+ ions is also 0.1000 M.
Using the formula, we can calculate the pH:
pH = -log(0.1000) = 1
So, the pH before any acid is added is 1.
(b) What is the pH when 25.00 ml of NaOH has been added?
To find the pH after NaOH has been added, we need to consider the neutralization reaction that occurs between NaOH and H+ ions produced by the dissociation of HOCl.
The balanced neutralization reaction is:
HOCl + NaOH → H2O + NaOCl
Since the molar ratio between HOCl and NaOH is 1:1, the number of moles of NaOH used will be equal to the number of moles of HOCl.
We have 25.00 ml of 0.1500 M NaOH, so we can calculate the moles of NaOH:
moles of NaOH = volume (L) × concentration (M)
= 0.02500 L × 0.1500 M
= 0.00375 moles of NaOH
Since the moles of HOCl is also 0.00375 (based on the 1:1 ratio), we can use the equation:
Molarity (M) = moles / volume (L)
To find the new concentration of HOCl:
new concentration of HOCl = moles of HOCl / volume (L)
= 0.00375 moles / 0.02500 L
= 0.150 M
Now, we can use the concentration of H+ ions (0.150 M) to calculate the new pH:
pH = -log(0.150)
= 0.824
So, the pH when 25.00 ml of NaOH has been added is 0.824.