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?

Answer 1

a) Henderson eq is pH = pka +log [A-]/[HA]   where HA is acid , A- is conjugate base

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

[A-] = 0.49 [HA]

b) A- moles = 0.49 HA moles

initially A- conc = HA con = 1M , let volume of A- be X liters , then HA volume is 1-X

then A- moles = M x V = 1X

Moles of HA = M x V = 1 x ( 1-X) = 1-X

from relation A_ moles = 0.49 HA moles we get

X = 0.49( 1-X)

X = 0.329 L   = 329 ml

hence A- solution volume required = 329 ml , HA solution volume = 1000-329 =671 ml

C) Initial HA moles = Mx V = 1 x 0.671 = 0.671 , A- moles = .329

NaOH moles = M x V = 2.3 x ( 29/1000) = 0.0667

NAOH gives OH- which reacts with HA to give A- and H2O

hence HA moles after reacting with OH- = 0.671-0.0667 =0.6043

A- moles now = 0.329+0.0667 = 0.3957

total volume = 1000+29 = 1029 ml = 1.029 L

pH = pka +log [A_]/[HA]

pH = 7.53 + log ( 0.3957 /1.029) // ( 0.6043/1.029)

    = 7.53 + log ( 0.3957/0.6043)

      = 7.35

Hence buffer was able to adequately handle addition of NaoH since pH change is not much

( old pH is 7.22 new pH is 7.35)