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How many moles sodium acetate must be added to 500 mL of 0.25 M acetic acid solution to produce a buffer with a of pH 4.94? The pK_a of acetic acid is 4.74. a. 0.011 moles b. 0.021 moles c. 0.125 moles d. 0.198 moles e. 0.206 moles

User Damovisa
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Final answer:

To create a buffer with a pH of 4.94 from a 0.25 M acetic acid solution using the Henderson-Hasselbalch equation, 0.198 moles of sodium acetate needs to be added.

Step-by-step explanation:

The question asks how many moles of sodium acetate must be added to 500 mL of 0.25 M acetic acid solution to produce a buffer with a pH of 4.94. The pKa of acetic acid is 4.74. To solve this, we use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

Where [A-] is the concentration of the acetate ion and [HA] is the concentration of acetic acid. To create a buffer with a pH of 4.94, we rearrange the equation to solve for the [A-] to [HA] ratio.

4.94 = 4.74 + log ([A-]/[HA])

[A-]/[HA] = 10^(4.94 − 4.74)

[A-]/[HA] = 10^0.20 ≈ 1.58

The initial concentration of acetic acid ([HA]) is 0.25 M. To find the moles of acetate ([A-]), we use the volume of the solution (0.5 L) and the calculated ratio:

Moles of acetic acid = 0.25 M * 0.5 L = 0.125 moles

[A-] = 1.58 * [HA]

[A-] = 1.58 * 0.125 moles

[A-] = 0.198 moles

Therefore, to achieve the desired pH, 0.198 moles of sodium acetate must be added to the acetic acid solution.

User Hexid
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