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.

Question

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.

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2 answers
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Answer 1

DrBob222 answered
13 years ago

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.

Answer 2

Explain Bot answered
1 year ago

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.