Question

50. ml of a 1.0 M solution of hydrochloric acid, HCl, is titrated with a 1.0 M solution of sodium hydroxide. What is the pH after 51 mL of NaOH has been added? Assume that the volumes are additive.

Answer 1

As the NaOH solution is added to the HCl solution, H⁺(aq) reacts with OH^–(aq) to form H₂O.

Both Na⁺ and Cl^– are spectator ions, having negligible effect on the pH.

In order to determine the pH of the solution, we must first determine how many moles of H⁺ were originally present and how many moles of OH^– were added. We can then calculate how many moles of each ion remain after the neutralization reaction.

In order to calculate [H⁺], and hence pH, we must also remember that the volume of the solution increases as we add titrant to the solution, thus diluting the concentration of all solutes present.

The number of moles of H⁺ in the original HCl solution = molarity * volume = 1.0 mol/litre * 0.050 litre = 0.05 moles

The number of moles of OH^– in 51.00 mL of 1 M NaOH = 1 * 0.051 = 0.051 moles

here we have crossed the equivalence point and the solution have more OH^- ions than H⁺ ions.

so here H⁺ is the limiting ion

=> 0.05 moles of H⁺ will react with 0.05 moles of OH^– ions

so the remaining moles of OH^– = 0.051 - 0.05 = 0.001 moles

During the course of the titration the volume of the reaction mixture increases as the NaOH solution is added to the HCl solution.

total volume of the solution = 50 + 51 = 101 ml = 0.101 L

hence the concentration of OH^– = moles / volume = 0.001 / 0.101 = 9.9*10^-3 moles / L

pOH of the solution = - log (concentration of OH^– ) = - log ( 9.9*10^-3 ) = 2.004 = 2

pH = 14 - pOH = 14 - 2 = 12