You prepare a 0.200M solution of a monoprotic buffer at pH 7.22. The pKa of this

Question

You prepare a 0.200M solution of a monoprotic buffer at pH 7.22. The pKa of this buffer is 7.53 at room temperature.

a. What is the molarity of the acid and conjugate base necessary to make the buffer.

[HA]=?

[A-]=?

b. A separate 1.0 M stock solution of the acid and conjugate base is made/ How many mL of each will you use to prepare 1.0 L of the buffer in part a?

mL of HA=?

mL of A-=?

c. You add 29mL of 2.3 M NaOH to your 1.0L of buffer. Is the buffer capacity able to adequately handle this new addition of NaOH? What will be your new pH?

Explanation / Answer

a) Let the acid be HA and the conjugate base be A-

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

=> 7.22 = 7.53 + log ([A-] / [HA])

=> log ([A-] / [HA]) = -0.31

=> ([A-] / [HA]) = 0.49

=> [A-] = 0.49 x [HA]

Given,

[HA] + [A-] = 0.2 M

=> [HA] + 0.49 x [HA] = 0.2

=> [HA] = 0.134 M = Acid

[A-] = 0.066 M = Conjugate Base

b) Moles of Acid in 1 L solution = 0.134 x 1 = 0.134 moles

Moles of conjugate base = 0.066 x 1 = 0.066 moles

Volume = Moles / Molarity

Volume of Acid solution = 0.134 / 1 = 0.134 L = 134 mL

Volume of Base solution = 0.066 / 1 = 0.066 L = 66 mL

c)

pH after adding NaOH

Moles of NaOH added = 2.3 x 0.029 = 0.0667 moles

pH = pKa + log (0.066 + 0.0667/ 0.134 - 0.0667)

=> pH = 7.53 + 0.295 = 7.825

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