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.

Since there is the same amount of agn03 in both silver chloride and silver nitrate they produce the same amount of moles. therefore silver chloride produces the same amount of silver nitrate, so your answer is 15.0 moles of silver nitrate. i hope this !