PH changes after acid/base additions to a buffer? ?

Question

I have a 100ml buffer comprised of 5ml 0.5M Na2HPO4 and 5mL 0.5M NaH2PO4. The pKa for this solution is 6.82. My question is, after adding 3mL of 1M HCl, how would I work out the new pH?

Using the Henderson-Hasselbalch equation is not an option, because if we begin with a ratio of 2.5:2.5 (the concentration of the base and acid respectively), with the addition of 3mM HCl, we end up with a theoretical concentration of the base which equals -0.05mM. Aside from the impossible negative concentration for my base, this figure also renders the Henderson-Hasselbalch equation useless, as it is not possible to take the log10 value of a negative number (that I'm familiar with).

Thanks, any help would be greatly appreciated!

Answer 1

Adding HCl will decrease the base component of the buffer (Na2HPO4) and increase the acid component (NaH2PO4) by the same amount. This change in moles equals the moles of the HCl. Get the new molarities or just use the adjusted moles to get buffer pH using Hendersen-Hasselbalch eqn.

Answer 2

ok well with the added HCl in excess of the buffer capacity this simply means that 2.5 mmmoles of the HCl will react with the 2.5 mmoles of the base form producing 2.5 MORE mmoles of the weak acid form...BUT leaving 0.5 mmoles of H+ NOT buffered .So this strong acid will inhibit any ionization of the weak acid and pH can be calculated directly from the excess strong acid....0.5 mmoles /103 mL = 4.85 X 10^-3 M

pH = -log 4.85 X 10^-3 = 2.32