Experiment #6 Acid/Base Titration 1. In a titration between HCl and NaOH, if the

Question

Experiment #6 Acid/Base Titration

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

2.In a titration, 13.61 mL of NaOH(aq) are titrated using 12.89 mL of 0.1077 M HCl(aq). Calculate the molarity of NaOH(aq).

(Notice that there is a (volume acid)/(volume base) ratio embedded in the calculation)

3.A 10.00 mL volume of vinegar (concentrated acetic acid solution) is diluted in a 100.0 mL volumetric flask to make a dilute acetic acid solution. The NaOH in #2 is used to titrate this dilute acetic acid. 14.53 mL of this diluted acetic acid solution are neutralized by 12.49 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 acid solution (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 acid solution (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 acetic acid (vinegar) The density of the vinegar is 1.011 g vinegar solution/mL vinegar solution.

(The percent Acetic Acid in Vinegar is usually 4% to 8% by mass)

Explanation / Answer

1. This will calculate more comcentration of NaOH using this formula M1V1= M2V2

2. M(NaOH)= M(HCl)*V(HCL) / V(NaOH)

= 0.1077*12.89/13.61 = 0.102002425M

3 b).M(acetic Aciid) = 0.102002425* 12.49/14.53 = 0.087681369 M

c) Concentration of Acetic Acid Solution = 10*0.087681369= 0.87681369 M

d) Mass of Acetic acid per litre = 0.87681369*60.052=52.654415712 gm/L

e)percentage = 52.654415712/1000 *100= 5.265 %

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