Calculate the H3O+ concentration at the halfway point when 34.5 mL of 0.19 M HBr is titrated with 0.1 M KOH. Assume additive

volumes.
Answer in units of M.

HBr + KOH ==> KBr + H2O

mols HBr = M x L = 0.0349 x 0.19 = 0.006555
mols KOH at halfway point = 1/2 x 0.006555 = 0.003278
mols HBr excess = 0.006555-0.003278 = 0.003278
M KOH = mols/L and L = mols/M = 0.003278/0.1 = 0.03278
total volume = 34.5 mL + 32,78 mL = ? and convert to L
Then M HBr = mol excesss HBr/total L = ?

to

Use the Henderson-Hasselbalch equation:

HBr + H2O ——> H3O^+ +Br^-

Ka=[H3O][Br^-]/[HBr]

-logKa=-log[H3O] -log[Br^-]/[HBr]

pKa=pH-log[Br^-]/[HBr]

pH=pKa+log[Br^-]/[HBr]

At the half equivalence point, HBr=Br^-, and thus log[1]=0

So, pH=pKa and pH=-log[H3O^+]

10^-(pKa)=H3O^+

Look in your text for the Ka value for HBr and solve for pKa, which is -log[Ka].

To anonymous:

HBr is a strong acid. The HH equation will not work unless you know the pKa and most likely that won't be listed anywhere. HH equation is for buffered solution and since HBr and KOH (both of them) are strong electrolytes, then KBr is the salt of a strong acid and a strong base and you don't have a buffered solution.

that was a bone head assumption: It is one of the strong acids, HBr that is. The setup will only work if the acid is a weak acid. So, the problem will require a little more work, which Dr. Bob222 essentially provided. All Halogen acids except for HF are strong acids. This is the first thing you learn when discussing acids and base strength in Chem 101.

I apologize

To calculate the H3O+ concentration at the halfway point, we need to determine the number of moles of HBr and KOH reacted.

Step 1: Calculate the number of moles of HBr.
First, we convert the volume of HBr solution to liters:
34.5 mL = 34.5 / 1000 = 0.0345 L

Next, we calculate the number of moles of HBr using the molarity:
moles of HBr = molarity × volume in liters
moles of HBr = 0.19 M × 0.0345 L
moles of HBr = 0.006855 mol

Step 2: Determine the number of moles of KOH reacted.
Since KOH reacts with HBr in a 1:1 stoichiometric ratio, the number of moles of KOH reacted is the same as the number of moles of HBr.

Step 3: Calculate the total volume of KOH solution required to reach the halfway point.
The halfway point occurs when half of the moles of HBr have reacted. Since the number of moles of KOH reacted is the same as the number of moles of HBr, the halfway point is reached when half of the initial moles of KOH have been used.

We can calculate the initial moles of KOH using the molarity and volume:
moles of KOH = molarity × volume in liters
moles of KOH = 0.1 M × volume in liters

At the halfway point, half of the initial moles of KOH have reacted, so:
0.5 × moles of KOH = moles of HBr
0.5 × (0.1 M × volume in liters) = 0.006855 mol

Now, solve for the volume of KOH solution required to reach the halfway point:
0.05 × (0.1 M × volume in liters) = 0.006855 mol
volume in liters = 0.006855 mol / (0.05 × 0.1 M)
volume in liters = 0.006855 mol / 0.005 M
volume in liters = 1.371 L

Step 4: Calculate the H3O+ concentration at the halfway point.
Since we assume additive volumes, the total volume at the halfway point is the sum of the initial volumes of HBr and KOH used:
total volume = HBr volume + KOH volume
total volume = 0.0345 L + 1.371 L
total volume = 1.4055 L

To find the H3O+ concentration at the halfway point, we divide the moles of HBr reacted by the total volume:
H3O+ concentration = moles of HBr / total volume
H3O+ concentration = 0.006855 mol / 1.4055 L
H3O+ concentration = 0.00488 M

Therefore, the H3O+ concentration at the halfway point is 0.00488 M.