A student titrates 0.025 M KOH into 50.00 ml of a solution of unknown weak acid HA. Equivalence point is reached when 30.00 ml of the KOH has been added. After 15.00ml of the KOH has been added the pH of the mixture is 3.92.
A. What is the concentration of HA ?
B. What is Ka for HA ?
C. What is the pH of the HA before any KOH is added?
D. What is the pH at equivalence point?
Do you know how to do any of this? If so post your work on what you know and I can help you through the remainder.
To solve this problem, we need to use the concepts of acid-base titration and the Henderson-Hasselbalch equation. Here's how we can approach each part of the question:
A. To determine the concentration of HA, we need to calculate the moles of KOH added at the equivalence point. We can then use this information to find the moles of HA. Finally, by dividing the moles of HA by the volume of the solution, we will obtain the concentration of HA.
1. Calculate the moles of KOH added at the equivalence point:
Molarity of KOH = 0.025 M
Volume of KOH at equivalence point = 30.00 ml = 0.030 L
Moles of KOH = Molarity × Volume = 0.025 M × 0.030 L
2. Determine the moles of HA:
According to the balanced equation for the reaction between KOH and HA:
1 mole of KOH reacts with 1 mole of HA
Therefore, the moles of HA will be equal to the moles of KOH.
3. Calculate the concentration of HA:
Concentration of HA = Moles of HA / Volume of solution
B. To find Ka for HA (the acid dissociation constant), we can use the pH at the halfway point of the titration (when 15.00 ml of KOH has been added). At this point, half of the HA will have reacted with KOH, resulting in a buffer solution.
1. Write the balanced equation for the reaction between HA and KOH. Since KOH is a strong base, it completely dissociates into K+ and OH- ions. The reaction can be represented as:
HA + OH- → A- + H2O
2. Determine the moles of HA and A- at the halfway point:
Moles of HA = Moles of KOH added / 2
Moles of A- = Moles of KOH added / 2
This is because the reaction between HA and KOH has a 1:1 stoichiometry.
3. Use the Henderson-Hasselbalch equation:
pKa = pH + log10([A-] / [HA])
Substitute the known values:
pKa = 3.92 + log10([A-] / [HA])
Since [A-] and [HA] are in a 1:1 ratio, the equation simplifies to:
pKa = 3.92 + log10(1) = 3.92
C. To determine the pH of HA before any KOH is added, we need to consider that initially, there is only HA present before the titration. This means that the solution consists of a weak acid, HA.
1. Set up the equilibrium expression for the dissociation of HA:
HA ⇌ H+ + A-
2. Since HA is a weak acid, we can assume that it only partially dissociates. Therefore, the initial concentration of HA will be equal to the concentration of the acid before any dissociation occurs.
3. Use the expression for the acid dissociation constant, Ka, to find the pH:
Ka = [H+][A-] / [HA]
Since initially, there are no H+ or A- ions in the solution, the equation simplifies to:
Ka = [H+]^2 / [HA]
Rearrange the equation to solve for [H+]:
[H+] = √(Ka × [HA])
4. Calculate the pH using the equation:
pH = -log10([H+])
D. To find the pH at the equivalence point, we need to consider the reaction between the strong base KOH and the weak acid HA at the stoichiometric point. At the equivalence point, all of the moles of HA will have reacted with KOH.
1. Use the balanced equation to determine the reaction between HA and KOH:
HA + KOH → K+ + A- + H2O
2. At the equivalence point, the moles of KOH added will be equal to the moles of HA initially present.
3. Calculate the volume of the solution at the equivalence point:
Volume of HA solution = 50.00 ml = 0.050 L
Volume of KOH at equivalence point = 30.00 ml = 0.030 L
4. Determine the concentration of OH- ions at the equivalence point:
Molarity of KOH = 0.025 M
Moles of KOH = Molarity × Volume = 0.025 M × 0.030 L
Moles of OH- = Moles of KOH / Volume of solution
5. Calculate the pOH using the equation:
pOH = -log10([OH-])
6. Find the pH using the equation:
pH = 14 - pOH
By following these steps, you should be able to answer each part of the question.