A 80.0 mL volume of 0.25 M HBr is titrated with 0.50M KOH. Calculate the pH after addition of 40.0 mL of KOH.
Saira, both GK and I told you how to solve this problem in an earlier post.
OMg iforgot i posted it up befoe sorry
To calculate the pH after the addition of 40.0 mL of 0.50 M KOH, we need to determine the amount of excess KOH and the resulting concentration of the remaining HBr.
First, let's calculate the moles of HBr initially present in the 80.0 mL of 0.25 M HBr solution:
Moles of HBr = Molarity × Volume
Moles of HBr = 0.25 M × 0.080 L
Moles of HBr = 0.020 mol
Since HBr and KOH react in a 1:1 ratio, the moles of KOH required to neutralize all the HBr can be determined by the stoichiometry.
Moles of KOH = Moles of HBr
Moles of KOH = 0.020 mol
Now, let's calculate the amount of excess KOH added (the moles of KOH in the 40.0 mL of 0.50 M KOH solution minus the moles of KOH required to neutralize the HBr):
Moles of excess KOH = (0.50 M)(0.040 L) - 0.020 mol
Next, we can calculate the total moles of KOH present:
Total moles of KOH = Moles of KOH initially + Moles of excess KOH
Total moles of KOH = 0.020 mol + [ (0.50 M)(0.040 L) - 0.020 mol ]
Now, let's calculate the total volume of the solution after the addition of KOH:
Total volume of the solution = Volume of HBr + Volume of KOH
Total volume of the solution = 80.0 mL + 40.0 mL = 120.0 mL = 0.120 L
Finally, we can calculate the concentration of HBr after the addition of KOH:
Concentration of HBr = Moles of HBr / Total volume of the solution
Concentration of HBr = 0.020 mol / 0.120 L
To find the pH, we first need to calculate the pOH, and then convert it to pH:
pOH = -log10(OH-) = -log10(MOH)
pOH = -log10(0.50 M) = -log10(0.50) = 0.301
Now, let's calculate the pH:
pH = 14 - pOH
pH = 14 - 0.301
pH ≈ 13.699
Therefore, the pH after the addition of 40.0 mL of 0.50 M KOH is approximately 13.699.
To calculate the pH after the addition of 40.0 mL of KOH, we need to determine the concentration of the resulting solution and then calculate the pH using the appropriate formula.
Step 1: Determine the moles of HBr in the initial solution.
Given:
Volume of HBr solution = 80.0 mL
Concentration of HBr solution = 0.25 M
Moles of HBr = Concentration × Volume
Moles of HBr = 0.25 mol/L × 0.080 L = 0.02 moles
Step 2: Determine the moles of KOH added.
Given:
Volume of KOH solution added = 40.0 mL
Concentration of KOH solution = 0.50 M
Moles of KOH = Concentration × Volume
Moles of KOH = 0.50 mol/L × 0.040 L = 0.02 moles
Step 3: Determine the moles of excess OH- ions.
Since the stoichiometric ratio between HBr and KOH is 1:1, the number of moles of excess OH- ions is equal to the number of moles of KOH added.
Moles of excess OH- ions = 0.02 moles
Step 4: Determine the remaining moles of HBr.
The moles of HBr initially present minus the moles of KOH added will give us the remaining moles of HBr.
Remaining moles of HBr = initial moles of HBr - moles of KOH added
Remaining moles of HBr = 0.02 moles - 0.02 moles = 0 moles
Step 5: Determine the total volume of the resulting solution.
The total volume of the resulting solution is the sum of the initial volume of HBr and the volume of KOH added.
Total volume = Volume of HBr solution + Volume of KOH solution added
Total volume = 80.0 mL + 40.0 mL = 120.0 mL
Step 6: Calculate the concentration of HBr in the resulting solution.
The concentration of HBr in the resulting solution can be determined by dividing the remaining moles of HBr by the total volume.
Concentration of HBr = Remaining moles of HBr / Total volume
Concentration of HBr = 0 moles / 0.120 L = 0 M
Since the concentration of HBr is now 0 M, the resulting solution will be neutral. Therefore, the pH of the resulting solution after adding 40.0 mL of KOH is 7.