A 23.0mL sample of a diprotic acid is titrated with 0.260M KOH. If 55.5mL of base is required to reach the second equivalence point, what is the concentration of the acid? I'm not sure how to do this.
H2A + KOH ==> KHA + H2O
KHA + KOH ==> K2A + H2O
If 55.5 mL KOH were required to titrate to the second equivalence, then 55.5/2 mL were required to reach the first equivalence point.
(55.5/2)*0.260 = ?? mmoles KOH.
mmoles H2A = the same.
M H2A = mmoles/mL
You know mmoles H2A and mL H2A, solve for M
7.2
To find the concentration of the diprotic acid, you need to use the concept of stoichiometry and the volume ratio between the acid and base in the titration.
First, let's break down the problem and identify the key information given:
- Volume of the diprotic acid (H2A) = 23.0 mL
- Concentration of the base (KOH) = 0.260 M
- Volume of the base (KOH) required to reach the second equivalence point (V2) = 55.5 mL
The second equivalence point occurs when all the moles of base required to neutralize both protons in the diprotic acid have reacted.
Now, let's use stoichiometry to find the number of moles of base required to reach the second equivalence point:
1. Calculate the number of moles of base (KOH) using the equation: moles = concentration (M) × volume (L)
Moles of KOH (n2) = 0.260 M × 0.0555 L
2. Since KOH is a monoprotic base and diprotic acid requires two moles of base for neutralization, we multiply the moles of KOH by 2:
Moles of H2A = 2 × Moles of KOH (n2)
3. Now that we have the moles of the diprotic acid (H2A) and the volume of the acid (V1), we can find its concentration (C1):
Moles of H2A = Concentration (C1) × Volume (L)
Concentration (C1) = Moles of H2A / Volume (L)
Concentration (C1) = (2 × Moles of KOH) / 0.023 L
Plug in the values calculated in steps 1 and 3 to find the concentration of the diprotic acid (H2A).