You are aiming to make a 250.0 mL buffer solution that is going to be at pH 4. U

Question

You are aiming to make a 250.0 mL buffer solution that is going to be at pH 4. Using the Henderson-Hasselback equation, calculate the molarity of [A-] that you will need in the buffer if it contains 75.0 mL of 1.50 M [HA]. The Ka of the HA is 2.35 x 10-5

Part C

You are provided with a stock solution of [A-] at 1.00 M. What is the volume of this solution that you need to add to the buffer you are aiming to make above.

Part D

What is the volume of deionized water that you will need to add to this [A-]/[HA] mixture that you just made to get the volume of the buffer (at pH 4) to 250.0 mL?

Explanation / Answer

Henderson-Hasselbalch equation:

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

   pH = -logKa + log ([A-]/[HA])

   4 = -log 2.35 x 10-5 + log ([A-]/[HA])

4 = - (-4.629) +  log ([A-]/[HA])

4 - 4.629 = log ([A-]/[HA])

-0.629 = log ([A-]/[75mL * 1.5mol/L]) = log ([A-]/[0.075L * 1.5mol/L]) = log ([A-]/0.1125mol)

[A-]/0.1125mol = 10-0.629 = 0.235 * 0.1125mol

[A-] = 0.0264 mol

molarity of [A-] = 0.0264 mol/250 mL = 0.0264 mol/0.25 L = 0.106 M

c)

volume = no. of moles / molarity

volume = 0.0264 moles / 1.0 mol/L = 0.0264 L = 26.4 mL

d) volume of deionized water = 250mL - (26.4mL + 75 mL)

   = 149 mL

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