Question

Use your mean NaOH molarity, the mean acid/base volume ratio from Project D, and the HCl/NaOH stoichiometric ratio from the balanced equation of Project D to calculate a precise concentration for (to standardize) your 0.1 M HCl. Show this as a single, sample calculation with explicit units in your notebook and on your report sheet. Given the information of mass of KHP is 0.9191g, Volume delivered is 45.03 mL, mean of molarity NaOH is 0.09971 mol/L, acid/base ration is 1.018, and chemical equation is HCl+NaOH -> NaCl+H2O. Please show work. x

Answer 1

You have found out the mean molarity of NaOH as 0.09971 mol/L from the standardization with KHP. Use this information and the balanced chemical equation.

NaOH (aq) + HCl (aq) -------> NaCl (aq) + H₂O (l)

The acid/base volume ratio is 1.018, i.e, 1.018 mL HCl reacts with 1.000 mL NaOH.

The volume of NaOH delivered is 45.03 mL; therefore, volume of HCl titrated = (45.03 mL NaOH)*(1.018 mL HCl/1.000 mL NaOH) = 45.84054 mL.

Again, the stoichiometric ratio of moles of NaOH and HCl is 1:1 as seen above.

Millimoles of NaOH = (volume of NaOH in mL)*(molarity of NaOH) = (45.03 mL)*(0.09971 mol/L) = 4.4899413 mmol.

As per the stoichiometric equation, millimoles of HCl titrated = millimoles of NaOH = 4.4899413 mmol.

Molarity of HCl = (millimoles of HCl)/(volume of HCl in mL) = (4.4899413 mmol)/(45.84054 mL) = 0.097947 mol/L ≈ 0.09795 mol/L (ans).