Question

1. In a titration between HCl and NaOH, if the volume of theNaOH used is larger than the volume of HCl used, would thiscalculate to a base concentration larger than or smaller than theacid concentration? Briefly explain.

2. In a titration, 13.51 mL of NaOH(aq) are titrated using 12.85mL of 0.1104 M HCl(aq). Calculate the molarity of NaOH(aq). (Noticethat there is a (volume acid)/(volume base) ratio embedded in thecalculation) !

3. A 10.00 mL volume of vinegar (concentrated acetic acidsolution) is diluted in a 100.0 mL volumetric flask to make adilute acetic acid solution. The NaOH in #2 is used to titrate thisdilute acetic acid. 14.53 mL of this diluted acetic acid solutionare neutralized by 12.51 mL NaOH of the molarity in #2. !

Calculate:

a. Volume Base/Volume Acid Ratio in the titration (4 s.f.)

b. Molarity of the diluted acetic acid using the ratio in a. (Ma= Mb*Ratio)

c. Molarity of the Acetic acid in the concentrated acetic acidsolution (vinegar) (Mc = Md*Vd/Vc where Vd = 100.0 mL and Vc =10.00 mL)

d. Mass of acetic acid per liter of the concentrated acetic acidsolution (vinegar) {the molar mass of acetic acid is 60.052 g/mol}(grams/L = mol/L * g/mol)

e. The percent by mass of acetic acid in the concentrated aceticacid (vinegar) The density of the vinegar is 1.011 g vinegarsolution/mL vinegar solution. (The percent Acetic Acid in Vinegaris usually 4% to 8% by mass)