Calculate the pH during the titration of 30.00 mL of 0.1000 M KOH with 0.1000 M HBr solution after the following additions of acid:
A.) 0 mL
I'm just trying to solve for A for now. It should be Ka=[H3O+][Br]/[HBr] but i dont have a Ka or a pKa.
Anonymous
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Since you are solving for the pH without the addition of HBr (strong acid), you are just finding out the initial conditions. Your titrant is HBr and your titrand is KOH. Thus, find the pH of KOH, a strong base. A strong base will dissociate completely as follows:
KOH --> K+ + OH-,
so if you have 0.1000 M KOH to begin, it will dissociate completely into 0.1000 M K and 0.1000 OH. From there you find the -log([OH]) to be 1, this is your pOH. Thus, your pH is just 14-pOH=13. Hope this helps!