Question

Textbook is showing HCl + NaOH titrating problem. After adding 5 mL of NaOH, why is the equation OH + H3O _--> 2h2o? Where is the h3o coming from

0.100MHC determihe the p ume of base required to reach the equivalence point, and then we during the titration. Consider the titration of 25.0 mL of Volume of NaOH Required to Reach the Equivalence Point tion, the added sodium hydroxide neutralizes the hydrochloric The equivalence point is reached when the number of moles of base added equals the ut its volume and ol ofac initial in solurtion. We calculate the amount of acid ini concentration: Initial mol HCl = 0.0250 L × 0.10-0.0 250 mol HCl The amount of NaOH that must be added is 0.00250 mol NaOH. We calculate the volume required from its concentration: of NH ルー= 0.0250 L 0.100 mol = 0.0250 L Volume NaOH solution = 0.00250 mol × The equivalence point is reached when 25.0mL of NaOH has been added. In this case, tions of both solutions are identical, so the volume of NaOH solution required to reach the point is equal to the volume of the HCI solution that is being titrated. the concerth equivlete Initial pH (Before Adding Any Base) The initial phH of the solution is simply the pld a 0.100 M HCI solution. Since HCl is a strong acid, the concentration of H,O" is also 0.100 M the pH is 1.00 pH-log[H,0"] -log(0.100) 1.00 pH After Adding 5.00 mL NaOH As NaOH is added to the solution, it neutralizes H,o OH (ag)+ HO (ag) 2 H,O() We calculate the amount of H,O at any given point (before the equivalence point) by using the rea- tion stoichiometry-1 mol of NaOH neutralizes 1 mol of HO*. The initial number of mols d H,o* (as we just calculated) is 0.00250 mol. We calculate the number of moles of NaOH addkd 5.00 mL by multiplying the added volume (in L) by the concentration of the NaOH solution: 0.100 mol IL mol NaOH added = 0.00500 L × =0.000500 mol NaOH The addition of OH causes the amount of H to decrease as shown in the following able OH (ag) HO (ag) 2 H2O(l) Before addition 0.00 mol 0.00250 mol