A50.0-ml volume of 0.15 m hbr is titrated with 0.25 m koh. calculate the ph after the addition of 11.0 ml of koh. express your answer numerically.

pH = 1.1 First, determine how many moles of each reactant was used Moles HBr = 0.0500 l * 0.15 mol/l = 0.0075 mol Moles KOH = 0.0110 l * 0.25 mol/l = 0.00275 mol Now determine how much acid (HBr) is left over after the rest of it is neutralized by the KOH. A simple matter of subtraction. 0.0075 - 0.00275 = 0.00475 mol Now determine the molarity of HBr. That would be moles per volume, so 0.00475 mol / (0.0500 l + 0.0110 l) = 0.00475 mol / 0.061 l = 0.077868852 mol/l = 0.077868852 m Now since HBr is a strong acid, it will completely dissociate into H+ and Br- ions. So the molarity of the H+ ions will be the same as the molarity of HBr. And the definition of pH is pH = - log(H+) So the log of 0.077868852 is -1.108636228 And the negation of that is 1.108636228 So the pH is 1.1 to 2 significant figures since that's the precision of the data we were given.