A buffer that is 0.271 M in acid, HA, and 0.150 M in the potassium salt of its conjugate base, KA, has a pH of 2.85. What is the pH of the buffer after 130. mL of 0.155 M LiOH is added to 0.650 L of this buffer? Assume that the volumes are additive.
First determine pKa.
pH = pKa + log(A^-)/(HA)
I have approximately 3 but you need to be more accurate than that.
650 mL x 0.271M = about 170 mmoles
650 mL x 0.150M = about 95 mmoles
mmoles LiOH added = 130 x 0.155M = about 20 mmoles.
...........HA + OH^- ==> A^- + H2O
initial...170....0........95
added...........20
change...-20....-20......+20
equil....150.....0........115
Substitute the equil values (after refining them) into the HH equation and solve for pH.
To find the pH of the buffer after adding LiOH, we need to calculate the concentration of the acid (HA) and its conjugate base (A-) in the final solution. Then, we can use the Henderson-Hasselbalch equation to determine the pH.
Step 1: Calculate the moles of acid and conjugate base in the original buffer solution.
Moles of HA = concentration of HA x volume of HA buffer solution
= 0.271 M x 0.650 L
Moles of KA = concentration of KA x volume of KA buffer solution
= 0.150 M x 0.650 L
Step 2: Calculate the moles of LiOH added.
Moles of LiOH = concentration of LiOH x volume of LiOH added
= 0.155 M x 0.130 L
Step 3: Determine the new volumes of the acid and conjugate base.
The moles of acid (HA) remain the same since its volume doesn't change. The volume of the conjugate base (A-) increases by the volume of LiOH added.
New volume of KA = volume of KA in buffer + volume of LiOH added
= 0.650 L + 0.130 L
Step 4: Calculate the new concentrations of acid and conjugate base.
New concentration of HA = moles of HA / new volume of HA
New concentration of A- = moles of KA / new volume of KA
Step 5: Determine the pH using the Henderson-Hasselbalch equation.
pH = pKa + log10([A-] / [HA])
The pKa is the negative logarithm of the acid dissociation constant (Ka) for the acid HA.
Now let's calculate the values:
Step 1: Moles of acid and conjugate base in the original buffer:
Moles of HA = 0.271 M x 0.650 L = 0.176 moles
Moles of KA = 0.150 M x 0.650 L = 0.098 moles
Step 2: Moles of LiOH added:
Moles of LiOH = 0.155 M x 0.130 L = 0.020 moles
Step 3: New volumes:
New volume of KA = 0.650 L + 0.130 L = 0.780 L
Step 4: New concentrations:
New concentration of HA = 0.176 moles / 0.650 L = 0.271 M
New concentration of A- = 0.098 moles / 0.780 L = 0.125 M
Step 5: pH calculation:
The pKa value is not provided in the question, so we cannot calculate the exact pH. The Henderson-Hasselbalch equation requires the pKa value. However, we can calculate the ratio [A-] / [HA] and use it to estimate the pH change.
[A-] / [HA] = (0.125 M) / (0.271 M) = 0.461
Now, we can estimate the pH change based on the ratio:
pH' = pH + log10(0.461)
Given that the initial pH is 2.85, we can calculate the final pH approximation:
pH' = 2.85 + log10(0.461)
pH' ≈ 2.85 - 0.336
pH' ≈ 2.514
Therefore, the pH of the buffer after adding LiOH is approximately 2.514.