In the titration of 238.0 mL of a 5.60×10-2 M solution of acid H3A (Ka1 = 1.0×10-3, Ka2 = 5.0×10-8, Ka3 = 2.0×10-12), calculate the volume of 2.90 M NaOH required to reach the following pH values. pH = 9.50 Volume required = mL pH = 4.70 Volume required = mL

When pH = 9.50 ; Volume = 14.8 L

When pH = 4.70 ; Volume = 4.51 L

Explanation:

We all know that:

So let us first calculate the pKa values of the following Ka:

Given that:

Ka1 = 1.0×10⁻³

Ka2 = 5.0×10⁻⁸

Ka3 = 2.0×10⁻¹²

For ; we have

=

= 3.00

=

= 7.30

=

= 11.70

At pH = 4.70:

The pH (4.70) is closer to than and .

However, the only important pKa for the dissociation of the acid will be directed towards only  

So: At pH = 4.70

-log = pH = 4.70

              ⇄            

where;

so, we have:

Given that;

283.0 mL of solution is given:

Then ; 283.0 mL = 0.283 L

However;

=

= 0.013328

=

= 0.01307 mole

moles of NaOH required to convert to  i.e

=

Finally; the volume of NaOH required =

= 0.004507

= 4.51 mL

When pH = 4.70 ; Volume = 4.51 L

When pH = 9.50

We try to understand that the average of and yields 9.50

i.e

=  

= 9.50

Here; virtually all,  is in  form;

SO; the moles of NaOH required to convert to will be:

= 2 × initial   = volume of NaOH × 1.8 M

= 2 × 1.3328 × 10⁻² mol = Volume  of NaOH × 1.8 M

Volume  of NaOH  =

= 0.1481 L

= 14.8 L

When pH = 9.50 ; Volume = 14.8 L