Question

A 50.0ml- volume of 0.15M HBr is titrated with 0.25M KOH . Calculate the pH after the addition of 13.0ml of KOH.

Answer 1

HBr is a strong acid therefore it will dissociate completely so whatever amount of moles of HBr there are is the initial amount of hydrogen ions that will be in solution.Same with KOH except it's a strong base.First find the mols of hydorgen and hydroxide.\[(.050L)\times(.15MHBr) = .0075molesH^+\] \[(.013L)\times(.25MKOH) = .00325moles OH ^{-}\] We can say that the moles of hydroxide will neutralize the H+ from HBr so that's just subtraction. We can also say that the conjugate base of HBr, a strong acid, will not effect ph. Therefore the concentration of H+ is the inital amount minus the OH- divided by the total volume.\[.0075 - .00325 = {.00425moles H^+ \over(.050L + .013L)}=.067M H^+\]Then convert to pH\[-\log[.067H^+]=1.171pH\]