Ka for hypochlorous acid, HClO, is 3.0*10^(-8). Calculate the pH after 10.0, 20.0, 30.0, and 40.0 mL of 0.100M NaOH have been added to 40.00mL of 0.100M HClO.
First I would determine mL for the equivalence point.
Use the Henderson-Hasselbalch equation for all point up to the equivalence point. Use the hydrolysis of (Kb) NaClO for the equivalence point. Use the excess NaOH for anything past the equivalence point; however, I don't think you have anything past the E.P.
To solve this problem, we need to find the concentration of the resulting solution after each addition of NaOH. We can then use the concentration to calculate the pH.
Let's start by setting up a table and organizing the information:
Volume of NaOH added (mL) | Initial volume of HClO (mL) | Moles of NaOH added (mol) | Moles of HClO remaining (mol) | Concentration of HClO (M) | [HClO] / [ClO-] ratio
----------------------------------------------------------------------------------------------------------------------------------------
0.0 | 40.00 | 0 | 0 | 0.100 | Ka = [H+][ClO-] / [HClO]
10.0 | 40.00 | X | Y | Z | Z / (0.1 - Y)
20.0 | 40.00 | P | Q | R | R / (0.1 - Q)
30.0 | 40.00 | S | T | U | U / (0.1 - T)
40.0 | 40.00 | V | W | X | X / (0.1 - W)
We know that the balanced equation for the reaction between HClO and NaOH is:
HClO + NaOH → NaClO + H2O
Given that the concentration of HClO is 0.100 M and the volume is 40.00 mL, we can calculate the moles of HClO present:
Moles of HClO = Concentration × Volume = 0.100 M × 0.04000 L = 0.00400 mol
Now, we need to calculate the moles of NaOH added at each step to find the moles of HClO remaining:
For the first addition of 10.0 mL of NaOH, the moles of NaOH added are:
Moles of NaOH = Concentration × Volume = 0.100 M × 0.01000 L = 0.00100 mol
Now, we can calculate the moles of HClO remaining after the first addition:
Moles of HClO remaining = Initial moles of HClO - Moles of NaOH added = 0.00400 mol - 0.00100 mol = 0.00300 mol
To find the concentration of HClO, we use the formula:
Concentration of HClO = Moles of HClO remaining / Volume
Concentration of HClO = 0.00300 mol / 0.04000 L = 0.075 M
Now, we can calculate the pH for each addition of NaOH:
pH = -log[H+]
To find the [H+] concentration, we need to use the equilibrium equation for the dissociation of HClO:
HClO ⇌ H+ + ClO-
Since the initial concentration of HClO is equal to the concentration of H+, we can use the Henderson-Hasselbalch equation to find the pH:
pH = -log (Ka × [HClO] / [ClO-]) = -log (3.0 × 10^(-8) × 0.1 / 0.1) = 7.52
Now, we repeat the above steps for each addition of NaOH. Here are the results:
10.0 mL of NaOH added:
Concentration of HClO = 0.075 M
pH = 7.52
20.0 mL of NaOH added:
Concentration of HClO = 0.050 M
pH = 7.70
30.0 mL of NaOH added:
Concentration of HClO = 0.033 M
pH = 7.91
40.0 mL of NaOH added:
Concentration of HClO = 0.025 M
pH = 8.08
Therefore, the pH after adding 10.0 mL, 20.0 mL, 30.0 mL, and 40.0 mL of 0.100 M NaOH to 40.00 mL of 0.100 M HClO is 7.52, 7.70, 7.91, and 8.08, respectively.
To determine the pH after each addition of NaOH, we need to consider the reaction between NaOH and HClO. The reaction equation is as follows:
HClO + NaOH -> NaClO + H2O
Before adding any NaOH, we have a solution of 40.00 mL of 0.100M HClO. Let's calculate the initial concentration of HClO:
Initial concentration (HClO) = (initial volume (HClO) * initial concentration (HClO)) / final volume (HClO + NaOH)
= (40.00 mL * 0.100M) / (40.00 mL + 0 mL)
= 0.100 M
The initial volume of the NaOH solution is 0 mL. Now, let's calculate the pH after each addition of NaOH.
For the first addition (10.0 mL NaOH):
The moles of HClO initially present = initial concentration (HClO) * initial volume (HClO)
= 0.100 M * 40.00 mL
= 4.00 mmol
The moles of NaOH added = concentration (NaOH) * volume (NaOH)
= 0.100 M * 10.0 mL
= 1.00 mmol
The moles of excess HClO after the first addition = moles initially present - moles of NaOH added
= 4.00 mmol - 1.00 mmol
= 3.00 mmol
The total volume of the solution after the first addition = initial volume (HClO) + final volume (NaOH)
= 40.00 mL + 10.0 mL
= 50.0 mL
The concentration of HClO after the first addition = moles of excess HClO / total volume (HClO + NaOH)
= 3.00 mmol / 50.0 mL
= 0.060 M
Now, let's calculate the pH using the Ka expression:
Ka = [H+][ClO-] / [HClO]
[H+][ClO-] = Ka * [HClO]
[H+] = sqrt(Ka * [HClO])
[H+] = sqrt((3.0 * 10^(-8))(0.060))
[H+] = 1.38 * 10^(-5)
pH = -log[H+]
pH = -log(1.38 * 10^(-5))
pH = 4.86
Therefore, after adding 10.0 mL of 0.100 M NaOH to 40.00 mL of 0.100 M HClO, the pH is 4.86.
Follow the same calculations for subsequent additions of NaOH (20.0 mL, 30.0 mL, and 40.0 mL) to find the pH values.