Question #1 Potasium hydroxide (known) volume used is 25 ml Molarity (concentration) = 0.150 M Moles of KOH used 0.150 × 25/1000 = 0.00375 moles Sulfuric acid (H2SO4) volume used = 15.0 ml unknown concentration The equation for the reaction is 2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l) Thus, the Mole ratio of KOH to H2SO4 is 2:1 Therefore, moles of H2SO4 used will be; 0.00375 × 1/2 = 0.001875 moles Acid (sulfuric acid) concentration 0.001875 moles × 1000/15 = 0.125 M
Question #2 Hydrogen bromide (acid) Volume used = 30 ml Concentration is 0.250 M Moles of HBr used; 0.25 × 30/1000 = 0.0075 moles Sodium Hydroxide (base) Volume used 20 ml Concentration (unknown) The equation for the reaction is NaOH + HBr = NaBr + H2O The mole ratio of NaOH : HBr is 1 : 1 Therefore, moles of NaOH used; = 0.0075 moles NaOH concentration will be = 0.0075 moles × 1000/20 = 0.375 M
