For 105ml of a buffer that is .40 M in HOCL and .36M in NaOCL what is the pH after 13.4 ml of 1.5M NaOH is added?
Ka for HOCL is 3.5x10^-8
105 mL x 0.4M = 42.0 millimoles HOCl
105 mL x 0.36M = 37.8 mmoles NaOCl.
13.4 mL x 1.5M NaOH = 20.1 mmoles NaOH.
.........HOCl + OH^- ==> OCl^- + H2O
Initial..42.0...0........37.8.....
add............20.1.................
change..-20.1..-20.1.....+20.1
equil.....?.....0.........?
Substitute the equil line into the Henderson-Hasselbalch equation and solve for pH.
To find the pH after adding NaOH to the buffer, we need to follow a few steps:
Step 1: Calculate the moles of HOCL and NaOCL in the initial buffer solution.
For HOCL:
moles of HOCL = volume of HOCL (in liters) * molarity of HOCL
= (105 ml / 1000) L * 0.40 M
= 0.042 mol
For NaOCL:
moles of NaOCL = volume of NaOCL (in liters) * molarity of NaOCL
= (105 ml / 1000) L * 0.36 M
= 0.038 mol
Step 2: Calculate the moles of NaOH added.
moles of NaOH = volume of NaOH (in liters) * molarity of NaOH
= (13.4 ml / 1000) L * 1.5 M
= 0.0201 mol
Step 3: Determine the limiting reagent.
Comparing the moles of NaOH (0.0201 mol) with the moles of HOCl (0.042 mol) and NaOCl (0.038 mol), we see that NaOH is the limiting reagent.
Step 4: Calculate the excess moles of NaOH.
excess moles of NaOH = moles of NaOH added - moles of HOCl
= 0.0201 mol - 0.042 mol
= -0.0219 mol (negative because it's in excess)
Step 5: Calculate the remaining moles of HOCl and NaOCl.
moles of HOCl remaining = moles of HOCl - excess moles of NaOH
= 0.042 mol - (-0.0219 mol)
= 0.0639 mol
moles of NaOCl remaining = moles of NaOCl - excess moles of NaOH
= 0.038 mol - (-0.0219 mol)
= 0.0599 mol
Step 6: Convert the remaining moles of HOCl and NaOCl to concentrations.
concentration of HOCl remaining = moles of HOCl remaining / volume of buffer (in liters)
= 0.0639 mol / (105 ml / 1000) L
= 0.608 M
concentration of NaOCl remaining = moles of NaOCl remaining / volume of buffer (in liters)
= 0.0599 mol / (105 ml / 1000) L
= 0.570 M
Step 7: Calculate the pKa from the given Ka value.
pKa = -log10(Ka)
= -log10(3.5x10^-8)
= 7.46
Step 8: Use the Henderson-Hasselbalch equation to calculate the pH.
pH = pKa + log10(concentration of NaOCl remaining / concentration of HOCl remaining)
= 7.46 + log10(0.570 / 0.608)
= 7.46 - 0.032
= 7.43
Therefore, the pH after adding 13.4 ml of 1.5 M NaOH to the buffer is approximately 7.43.