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.
To find the pH at each volume of base added, we can use the Henderson-Hasselbalch equation, which relates the pH of a solution to the pKa of the acid and the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given as:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the negative logarithm of the acid dissociation constant (Ka) of the weak acid
[A-] = concentration of the conjugate base
[HA] = concentration of the weak acid
In this case, the weak acid is HA and its conjugate base is A-. Initially, before any base is added, the concentration of the weak acid [HA] is 0.100 M and the concentration of the conjugate base [A-] is 0 M.
Given the pKa of 5.00, we can use this equation to find the pH at each volume of base added.
At Vb = 0 mL:
- [HA] = 0.100 M
- [A-] = 0 M
- pH = pKa + log([A-]/[HA])
= 5.00 + log(0/0.100)
= 5.00 (since log(0) is undefined)
- Therefore, the pH is 5.00
At Vb = 1 mL:
- [HA] = 0.100 M
- [A-] = volume of base added * Molarity of the base / (volume of acid solution + volume of base added)
= 1 mL * 1.00 M / (100.0 mL + 1 mL)
≈ 0.01 M
- pH = pKa + log([A-]/[HA])
= 5.00 + log(0.01/0.100)
≈ 4.00
- Therefore, the pH is approximately 4.00
By following a similar procedure, you can calculate the pH at each subsequent volume of base added (Vb = 5 mL, 9 mL, 9.9 mL, 10 mL, 10.1 mL, and 12 mL). Simply substitute the appropriate values into the Henderson-Hasselbalch equation.
To create a graph of pH versus Vb, plot the corresponding pH values on the y-axis and the volumes of base added (Vb) on the x-axis, using the data calculated above. Connect the points with a line to visualize the trend of pH as the volume of base added increases.