Calculate pH of Buffer Solution & Understand Ka Equilibrium

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In summary, a student measures out 27 cm3 of 0.050 mol dm-3 HA solution and reacts with one-third of the volume of 0.10 mol dm-3 sodium hydroxide needed for complete neutralisation. The volume of sodium hydroxide used is 4.5 cm3 and the pH of the resulting buffer solution is 2.73, with a Ka value of 9.3 x 10-4. The ratio of [A-]/[HA] is 0.5, determined by neutralizing 1/3 of the acid and resulting in a solution with one part A- and two parts HA.
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Homework Statement


A student sets out to make a buffer solution. The student measures out 27 cm3 of 0.050 mol dm-3 HA solution and reacts with one-third of the volume of 0.10 mol dm-3 sodium hydroxide needed for complete neutralisation.

(i) Calculate the volume of sodium hydroxide that the student uses.

(ii) Calculate the pH of the resulting buffer solution, given that Ka = 9.3 x 10-4.


Homework Equations


n = cV
HA <--> H+ + A- (<--> indicating a dynamic equilibrium)


The Attempt at a Solution



For (i) I'm getting 4.5 cm3 which is in agreement with the mark scheme.

For (ii) the answer is 2.73, but I am not getting that, I'm getting 3.33.

Their answer implies that [A-]/[HA] = 0.5, but how are they getting this?
 
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  • #2
If you neutralize 1/3 of the acid, solution contains one part of A- and two parts of HA, hence [A-]/[HA] = 1/2.

Actually exact answer is neither 2.73 nor 3.33 :wink:
 
  • #3
Thanks
 

1. What is a buffer solution?

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added to it. It is typically made up of a weak acid and its conjugate base, or a weak base and its conjugate acid.

2. How do you calculate the pH of a buffer solution?

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation, which is pH = pKa + log([base]/[acid]), where pKa is the negative log of the acid dissociation constant (Ka) and [base] and [acid] are the concentrations of the base and acid in the buffer solution.

3. What is the relationship between Ka and pKa?

Ka and pKa are related by the equation pKa = -log(Ka). Ka is a measure of the strength of an acid, where a higher Ka indicates a stronger acid. pKa is the negative logarithm of Ka and is used to compare the strengths of different acids. The lower the pKa, the stronger the acid.

4. How does equilibrium play a role in buffer solutions?

Buffer solutions rely on the equilibrium between a weak acid and its conjugate base (or a weak base and its conjugate acid) to maintain a constant pH. When small amounts of acid or base are added to the buffer solution, the equilibrium shifts to either the left or the right, minimizing the change in pH.

5. How do you use a buffer solution in a laboratory setting?

Buffer solutions are commonly used in laboratory settings to maintain a specific pH during experiments. They are especially useful when working with sensitive biological systems that require a specific pH range. Buffers can also be used to calibrate pH meters and to maintain a constant pH in chemical reactions.

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