Question

Determine the volume (in mL) of 1.00mol/L HCl that must be added to 750mL of 0.50mol/L HPO₄^2- to produce a buffer with a pH of 7.00.

Answer 1

The buffer equation (henderson haselbach) is given by

pH = pKa + log(A-/HA)

where

A- = conjugate base, in this case is the HPO42-

HA = acid (formed when HCl is added), H2PO4-

so

H2PO4- <--> H+ + HPO42-

this is the second ionizaiton point, meaning that we are interested in the 2nd Ka point

pKa2 = 7.21

so...

pH = 7.21+ log(A-/HA)

now, we need to find A- and HA, in equilibirum, i.e. HPO4-2 and H2PO4-

initially:

mmol of HPO4-2 = MV = 750*0.50 = 375

mmol of H2PO4- = MV = 0

after addition of Vacid of M = 1 M

mmol of acid = MV = 1*Vacid

reaction:

mmol of HPO4-2 left = 375 - 1*Vacid

mmol of H2PO4- formed = MV = 0 +1*Vacid

then, we can substitute this in the pH equation.

pH = 7.21+ log(A-/HA)

7.0= 7.21+ log(A-/HA)

7.0= 7.21+ log((375 - 1*Vacid)/(1*Vacid))

solve for Vacid:

10^(7-7.21) = ((375 - 1*Vacid)/(1*Vacid))

0.6165* Vacid = 375 - Vacid

1.6165*Vacid = 375

Vacid = 375/1.6165 = 231.98 mL of acid required to get a pH = 7