A 30.00 mL volume of a weak acid, HA, (Ka = 3.8 x 10^-6) is titrated with 39.00 mL of 0.0958 M NaOH to the equivalence point.
A.) What is the pH of the acid solution, before any base is added to it?
B.) What is the pH of the acid solution after exactly 19.50 mL of NaOH is added?
HA ==> H^+ + A^-
Ka = (H^+)(A^-)/(HA)
A) pure HA.
Set up ICE chart for pure HA, substitute into Ka expression, solve for (H^+) and convert to pH.
B)
moles HA initially = M x L = ??
moles NaOH at 19.5 mL = M x L = ??
Subtract mols HA - moles NaOH. The difference is the amount of HA remaining, moles NaOH is the amount of A^- formed. Convert moles of HA and A^- to molarity, plug into Ka expression, and solve for H^+, then convert to pH. As an alternative, you may plug those concns into the Henderson-Hasselbalch equation and obtain pH directly.
To answer these questions, we need to consider the stoichiometry and the equilibrium of the acid-base reaction.
A.) To find the pH of the acid solution before any base is added, we need to determine the concentration of the acid solution.
Given:
- Volume of HA = 30.00 mL
- Ka (acid dissociation constant) of HA = 3.8 x 10^-6
1. Convert the volume of HA to liters:
Volume of HA = 30.00 mL = 0.03000 L
2. Use the equation for Ka to set up an expression for the concentration of HA (let's call it [HA]) in terms of the initial volume and Ka:
Ka = [H+][A-]/[HA]
Since [H+] = [A-] (as it is a weak acid), we can write:
Ka = [H+][H+]/[HA]
3. Rearrange the equation to solve for [HA]:
[HA] = [H+]²/Ka
4. Substitute the value of Ka and solve for [HA]:
[HA] = (x)²/3.8 x 10^-6
where x represents the concentration of [H+] or [A-].
Since the volume of HA solution is given, the initial concentration of HA can be calculated using the following equation:
Molarity (M) = moles/volume
5. Calculate the initial moles of HA:
Moles of HA = Molarity (M) x Volume
Since the volume is given as 30 mL = 0.03 L, we can plug in the values:
Moles of HA = Molarity (M) x 0.03 L
B.) To find the pH of the acid solution after exactly 19.50 mL of NaOH is added, we need to determine the number of moles of acid that react with the base.
Given:
- Volume of NaOH = 39.00 mL
- Concentration of NaOH = 0.0958 M
- Volume of NaOH added = 19.50 mL
1. Convert the volumes to liters:
Volume of NaOH = 39.00 mL = 0.03900 L
Volume of NaOH added = 19.50 mL = 0.01950 L
2. Calculate the moles of NaOH added:
Moles of NaOH = Molarity (M) x Volume
Moles of NaOH = 0.0958 M x 0.01950 L
3. Use the stoichiometry of the reaction to determine the moles of HA that reacted with NaOH:
Based on the balanced equation for the reaction:
HA + NaOH → NaA + H2O
1 mole of HA reacts with 1 mole of NaOH
Therefore, the moles of HA that reacted with NaOH is equal to the moles of NaOH added.
Once you have the moles of HA that reacted with NaOH, you can recalculate the new concentration of HA and then determine the pH using the same methodology as in part A.
Keep in mind that the change in concentration of [HA] will affect the concentration of [H+] and hence the pH. So, you need to recalculate the new concentration of HA after the addition of NaOH to find the new pH.