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)

pH = pka + log(A-/HA)

   7.22 = 7.53+log(x)

x= A-/HA = 0.5

HA + A- = 0.2 M

HA = 0.2/1.5 = 0.133 M

A- = 0.2-0.133 = 0.067 M

b)

pH = pka + log(A-/HA)

   7.22 = 7.53+log(x)

x= A-/HA = 0.5

HA + A- = 0.2 mol

HA = 0.2/1.5 = 0.133 mol

Volume of HA = 0.133/1 = 0.133 L = 133 ml

A- = 0.2-0.133 = 0.067 mol = 0.067 ml

for 1 litre solution take

Volume of HA = 133*5 = 665 ml

Volume of A- = 67*5 = 335 ml

c.

No of mol of NaOH = 29/1000*2.3 =   0.0667 mol

No of mol of HA = 0.133*5 = 0.665 mol

No of mol of A- = 0.067*5 = 0.335 mol

pH = 7.53+log((0.335+0.067)/(0.665-0.067))

   = 7.35