Ka for hypochlorous acid,HClO is 3.0x10^-8. Calculate the pH after 10.0, 20.0, 30.0, and 40.0mL of 0.100M NaOH have been added to 40.0mL of 0.100M HOCl.
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To solve this problem, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka)
[A-] = the concentration of the conjugate base
[HA] = the concentration of the acid
In this case, the acid is HOCl (hypochlorous acid), and the conjugate base is OCl- (hypochlorite ion).
Given:
Ka = 3.0x10^-8
Volume of HOCl = 40.0 mL
Concentration of HOCl = 0.100 M
Volume of NaOH added = 10.0, 20.0, 30.0, and 40.0 mL
Concentration of NaOH = 0.100 M
Step 1: Calculate the number of moles of HOCl initially present:
moles of HOCl = concentration of HOCl × volume of HOCl
moles of HOCl = 0.100 M × 0.040 L (convert mL to L)
moles of HOCl = 0.004 mol
Step 2: Calculate the number of moles of NaOH added:
moles of NaOH = concentration of NaOH × volume of NaOH added
moles of NaOH = 0.100 M × (10.0 mL + 20.0 mL + 30.0 mL + 40.0 mL) (convert mL to L)
moles of NaOH = 0.100 M × 0.100 L
moles of NaOH = 0.010 mol
Step 3: Determine the excess of NaOH:
moles of excess NaOH = moles of NaOH - moles of HOCl
moles of excess NaOH = 0.010 mol - 0.004 mol
moles of excess NaOH = 0.006 mol
Step 4: Calculate the final volume of the solution:
final volume = volume of HOCl + volume of NaOH added
final volume = 0.040 L + (10.0 mL + 20.0 mL + 30.0 mL + 40.0 mL) (convert mL to L)
final volume = 0.040 L + 0.100 L
final volume = 0.140 L
Step 5: Calculate the concentration of HOCl and OCl- in the final solution:
concentration of HOCl = moles of HOCl / final volume
concentration of HOCl = 0.004 mol / 0.140 L
concentration of HOCl = 0.0286 M
concentration of OCl- = moles of excess NaOH / final volume
concentration of OCl- = 0.006 mol / 0.140 L
concentration of OCl- = 0.0429 M
Step 6: Calculate the pH using the Henderson-Hasselbalch equation for each case:
For 10.0 mL of NaOH added:
pH = pKa + log([OCl-]/[HOCl])
= -log(3.0x10^-8) + log(0.0429 M / 0.0286 M)
For 20.0 mL of NaOH added:
pH = pKa + log([OCl-]/[HOCl])
= -log(3.0x10^-8) + log(0.0857 M / 0.0286 M)
For 30.0 mL of NaOH added:
pH = pKa + log([OCl-]/[HOCl])
= -log(3.0x10^-8) + log(0.1286 M / 0.0286 M)
For 40.0 mL of NaOH added:
pH = pKa + log([OCl-]/[HOCl])
= -log(3.0x10^-8) + log(0.1714 M / 0.0286 M)
Now, calculate each pH value accordingly and round to the desired number of decimal places.
To solve this problem, we can use the concept of titration and the Henderson-Hasselbalch equation. Let's break down the steps to find the pH after each addition of NaOH:
Step 1: Write the balanced equation for the reaction between NaOH and HOCl:
HOCl + NaOH → NaOCl + H2O
Step 2: Convert the volumes of the solutions to moles:
For the initial solution (before any NaOH is added):
The moles of HOCl = (0.040 L) x (0.100 mol/L) = 0.004 mol
For the NaOH solutions added:
The moles of NaOH = (0.010 L) x (0.100 mol/L) = 0.001 mol
Step 3: Calculate the moles of reactants and products remaining after each addition of NaOH.
Since the balanced equation has a 1:1 ratio of HOCl to NaOH, the number of moles of HOCl remaining will decrease by the number of moles of NaOH added in each step.
Step 4: Calculate the concentration of the remaining HOCl after each addition.
The total volume after each addition of NaOH will be 40.0 mL + 10.0 mL, 20.0 mL, 30.0, or 40.0 mL.
Step 5: Use the Henderson-Hasselbalch equation to calculate the pH.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka) = -log(Ka)
[A-] = the concentration of the conjugate base (OCl-)
[HA] = the concentration of the acid (HOCl)
Step 6: Calculate the pH using the Henderson-Hasselbalch equation for the remaining HOCl in each step after adding NaOH.
Let's calculate the pH after each addition of NaOH:
After adding 10.0 mL of NaOH:
Moles of HOCl remaining = 0.004 mol - 0.001 mol = 0.003 mol
Total volume = 40.0 mL + 10.0 mL = 50.0 mL = 0.050 L
[OCl-] = 0.001 mol / 0.050 L = 0.020 M
[HOCl] = 0.003 mol / 0.050 L = 0.060 M
pH = pKa + log([OCl-]/[HOCl])
= -log(3.0x10^-8) + log(0.020/0.060)
= 7.52
After adding 20.0 mL of NaOH:
Moles of HOCl remaining = 0.003 mol - 0.001 mol = 0.002 mol
Total volume = 40.0 mL + 20.0 mL = 60.0 mL = 0.060 L
[OCl-] = 0.002 mol / 0.060 L = 0.033 M
[HOCl] = 0.002 mol / 0.060 L = 0.033 M
pH = pKa + log([OCl-]/[HOCl])
= -log(3.0x10^-8) + log(0.033/0.033)
= 7.00
You can follow a similar process to find the pH after adding 30.0 mL and 40.0 mL of NaOH.