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^+
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.
5 answers
Look in your text for the Ka value for HBr and solve for pKa, which is -log[Ka].
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
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
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.
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
I apologize