1.6g of an unknown monoprotic acid (HA) required 5.79 mL of a 0.35 M NaOH solution to reach the equivalence point. Determine the molar mass of the acid and given that the pH at the half way point to the equivalent point is 3.86. Calculate the Ka of the unknown acid.
molar mass= 90.01 g/mol and Ka= 1.38 x 10^-4
Na(oh)3
To determine the molar mass of the acid (HA):
1. Start by calculating the number of moles of NaOH used to reach the equivalence point.
- Moles of NaOH = concentration of NaOH (in M) × volume of NaOH (in L)
- Moles of NaOH = 0.35 M × 0.00579 L
2. Since the monoprotic acid (HA) reacts in a 1:1 ratio with NaOH, the number of moles of NaOH used is equal to the number of moles of HA present in the solution.
- Moles of HA = moles of NaOH = 0.35 M × 0.00579 L
3. Next, calculate the molar mass of HA using the mass of HA given and the number of moles you just calculated.
- Molar mass of HA (in g/mol) = mass of HA (in g) / moles of HA
4. Plug in the values to calculate the molar mass of HA:
- Molar mass of HA = 1.6 g / (0.35 M × 0.00579 L)
Now, let's move on to calculating the Ka of the unknown acid:
5. The pH at the halfway point to the equivalence point (which is also the half-equivalence point) is given as 3.86. At the half-equivalence point, half of the acid has reacted with an equal amount of the base, forming a buffer solution. The pH of a buffer is related to the pKa of the acid component by the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
At the half-equivalence point, [A-] = [HA], because they are in equal amounts. Thus, the equation becomes:
pH = pKa + log (1)
Since log (1) is 0, the equation simplifies to:
pH = pKa
Therefore, the pKa at the half-equivalence point is 3.86.
6. Convert the pKa to Ka using the equation:
Ka = 10^(-pKa)
Ka = 10^(-3.86)
Plug-in the values to calculate the Ka of the unknown acid.
HA + NaOH ==> NaA + H2O
mols NaOH = M x L = ?
mols HA = mols NaOH (look at the coefficient in the balanced equation).
mols HA = grams HA/molar mass HA. You know mols and grams, solve for molar mass.
If the pH at the half way point is 3.86, then the pKa is 3.86.
pKa = -log Ka. Solve for Ka.