A Compound (not specified) has 2 ionizable groups. One is at pKa 4.2 and the other pKa is between 7 and 10. If you have 10.0 mL of a 1.0M solution at a pH of 8.17 and you add 10.0mL of 1.0M HCl which changes the pH to 4.2 what is the second pKa? I tried working this problem out a few different ways and keep getting different answers. The last calculation I tried gave me a pKa of 4.44. I'm using pH = pKa + log [A-]/[HA] I used pH = pKa-log[HA] / 2 to get an HA of 1.38. I plugged 8.17 and 4.2 in respectively as pH and pKa. Looking at it now that doesn't seem right. But where I get stuck is how to add the 0.01 moles of HCl into the equation. I also was wondering if the answer is much more simple than I think it is. At the second ionizable group's pKa it would equal the pH right? So if that's the case is the pKa just 8.17? Or am I just completely wrong?