pls help.
A weak acid HA (pKa = 5.00) was titrated with 1.00 M KOH. The acid solution had a volume of 100.0 mL and a molarity of 0.100 M. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL.
Well, well, well! Looks like we've got a titration question here. Don't worry, I've got the perfect pH recipe for you. Let's clown around with some math, shall we?
To find the pH at each volume of base added, we need to use the Henderson-Hasselbalch equation, which goes like this:
pH = pKa + log([A-]/[HA])
Where [A-] is the concentration of the conjugate base and [HA] is the concentration of the weak acid.
Now, let's grab our clown calculator and dive right in!
When Vb = 0 mL:
Since no base has been added, we're solely dealing with the weak acid. Calculate the concentration of HA using the initial volume and molarity. Then plug the values into the Henderson-Hasselbalch equation and see what pH you get.
When Vb = 1 mL:
Now that we've added 1 mL of base, we'll have a new volume of acid and base to consider. Calculate the new concentrations of HA and A-, and use the Henderson-Hasselbalch equation once more.
Repeat this process for each volume of base added, and voilà! You'll have a juicy graph of pH versus Vb that'll make all the other chemists laugh with glee.
Just remember to be cautious around acid and base solutions - they're no joking matter! Safety first, always.
I hope this explanation puts a smile on your face. Happy titrating!
To find the pH at different volumes of base added, we need to understand the process of titration and the relationship between the volume of base added and the resulting pH.
First, let's analyze the given information:
- Weak acid HA with pKa = 5.00
- Concentration of acid solution (HA) = 0.100 M
- Volume of acid solution (HA) = 100.0 mL
- Concentration of KOH (base) = 1.00 M
- Volumes of base added: Vb = 0, 1, 5, 9, 9.9, 10, 10.1, and 12 mL
To find the pH at each volume, we will use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa and the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-] / [HA])
Since the concentration of HA initially is given and the volume changes as the base is added, we need to calculate the concentrations of the acid (HA) and its conjugate base (A-) at each volume.
Let's calculate the pH at each volume of base added:
1. Vb = 0 mL:
At this stage, no base has been added. Therefore, the concentration of acid (HA) remains the same.
[HA] = 0.100 M
[A-] = 0 M (no conjugate base formed yet)
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
pH = 5.00 + log (0 / 0.100)
pH = 5.00 - ∞ (log of 0 is undefined)
The pH is undefined when no base has been added because there is no conjugate base present to balance the acid.
2. Vb = 1 mL:
At this stage, we add 1 mL of the base. To calculate the concentrations, we use the balanced chemical equation to determine the mole ratio between the acid and base. In this case, it is 1:1.
Since we are using 1.00 M KOH, the concentration of OH- added (A-) will be 1.00 M at each volume of base added.
[HA] = initial concentration - (volume of base added * concentration of base)
[HA] = 0.100 M - (0.001 L * 1.00 M)
[HA] = 0.100 M - 0.001 M
[HA] = 0.099 M
Using the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
pH = 5.00 + log (1.00 / 0.099)
pH ≈ 5.70
So, at Vb = 1 mL, the pH is approximately 5.70.
3. Repeat the calculations for each volume of base added (5, 9, 9.9, 10, 10.1, and 12 mL) using the Henderson-Hasselbalch equation.
Finally, plot the graph of pH vs. Vb using the volumes of base added and their corresponding pH values.
To determine the pH at the given volumes of base added, we need to consider the titration of the weak acid HA with the strong base KOH.
Here's a step-by-step approach to finding the pH at each volume of base added:
1. Calculate the initial moles of acid (HA) present:
Moles of HA = Volume of acid solution (in L) x Molarity of acid (in mol/L)
In this case, the volume of the acid solution is given as 100.0 mL, which is equivalent to 0.100 L. The molarity of the acid is 0.100 M.
Therefore, Moles of HA = 0.100 L x 0.100 mol/L = 0.010 mol
2. Determine the volume of base (KOH) required to reach each given volume of base added:
The volume of base added (Vb) represents the amount of KOH required to neutralize the acid.
To find the volume of base required to reach each Vb, we use the equation: Moles of KOH = Moles of HA
Rearrange the equation to solve for the volume of KOH required:
Volume of KOH = Moles of KOH / Molarity of KOH
Since the molarity of KOH is 1.00 M, we can use the equation Volume of KOH = Moles of KOH / 1.00 M.
Substitute the value of Moles of KOH with the calculated moles of HA from step 1.
3. Calculate the moles of base (KOH) added at each volume (Vb) of base added:
To calculate the moles of base added at each volume, we use the equation Moles of KOH = Volume of KOH x Molarity of KOH.
4. Calculate the remaining moles of acid (HA) at each volume (Vb) of base added:
The remaining moles of acid can be determined using the equation Moles of HA remaining = Initial moles of HA - Moles of KOH added.
5. Calculate the concentration of acid (HA) remaining at each volume (Vb) of base added:
Concentration of HA remaining = Moles of HA remaining / Total volume (sum of acid volume and base volume)
6. Calculate the pOH at each volume (Vb) of base added:
The pOH can be determined using the equation pOH = -log10 (concentration of HA remaining)
7. Calculate the pH at each volume (Vb) of base added:
To calculate the pH, we need to consider the ionization of water and the autoionization constant (Kw = [H+][OH-]).
pH = 14 - pOH
Using the above steps, perform the calculations for each given volume of base added to find the corresponding pH values. Once you have the pH values, plot a graph of pH versus Vb by representing the given volumes on the x-axis and the calculated pH values on the y-axis.
1. First calculate the volume KOH needed to reach the equivalcnce point. That will tell you if the volume(s) in the problem are BEFORE or AFTER the equivalence point.
HA + KOH ==> KA + H2O
2. For the beginning, zero mL.
.........HA ==> H^+ + A^-
I........0.1....0......0
C.........-x....x......x
E.......0.1-x....x......x
Ka = (H^+)(A^-)/(HA)
Substitute from the E line and solve for x = (H^+) then convert to pH.
b. All points before the eq. pt. I will use 5 mL KOH as an example.
100 mL HA x 0.1M = 10 millimoles
5 mL KOH x 1M = 5 mmoles KOH.
10-5 = 5 mmols HA remains.
5 mmols KA formed. (HA) = 5 mmols/105 mL = ?
(A^-) = 5 mmols/105 mL = ?
Substitute into the Ka expression and solve for (H^+), then convert to pH.
c. At the equivalence point the solution is KA and the A^- is a base.
(KA) = (100 x 0.1)/110 mL = about 0.09M
..........A^- + HOH ==> HA^- + OH^-
I.......0.09............0......0
C.........-x............x......x
E.......0.09-x..........x.......x
Kb for A^- = (Kw/Ka for HA) = (x)(x)/(0.09-x)
Substitute and solve for x = (OH^-), then convert to pH.
d. All points after the eq pt.
mmoles HA initially.
mmols KOH added.
KOH will be in excess; therefore,
mmoles KOH-mmoles HA = mmols KOH in excess. M KOH = mmol/mL and that gives you (OH^-). Then convert to pH.