1. In a titration between HCl and NaOH, if the volume of the NaOH used is larger
ID: 835719 • Letter: 1
Question
1. In a titration between HCl and NaOH, if the volume of the NaOH used is larger than the volume of HCl used, would this calculate to a base concentration larger than or smaller than the acid concentration? Briefly explain.
2. In a titration, 13.51 mL of NaOH(aq) are titrated using 12.85 mL of 0.1104 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 12.51 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 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. In a titration between HCl and NaOH, if the volume of the NaOH used is larger than the volume of HCl used, would this calculate to a base concentration larger than or smaller than the acid concentration? Briefly explain.
ans)
NaOH +HCl ---> NaCl + H2O
so,
m1v1=m2v2
so if more volume of NaOH has been added, the volume of molarity of NaOH reported would be less since the number of moles is being divided by higher volume thus giving the result to be of low value.
2. In a titration, 13.51 mL of NaOH(aq) are titrated using 12.85 mL of 0.1104 M HCl(aq). Calculate the molarity of NaOH(aq). (Notice that there is a (volume acid)/(volume base) ratio embedded in the calculation)
ans)
let the molarity of NaOH be v M.so equating the number of moles,
12.85*0.1104=13.51*v
or v=0.105 M
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 12.51 mL NaOH of the molarity in #2.
Calculate:
a. Volume Base/Volume Acid Ratio in the titration (4 s.f.)
volume base/volume cid=12.51/14.53
=0.8609
b. Molarity of the diluted acetic acid using the ratio in a. (Ma = Mb*Ratio)
molarity of the dilute acid=0.8609*0.105
=0.0904 M
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)
molarity of the concentrated acid be v.so,
10*v/100=0.0909 M
or v=0.904 M
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)
mass of ecid per liter of the solution=0.904*60.052
=54.3 grams
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)
mass of the solution of 1 liter=1011 grams
so percent of acetic acid=54.3*100/1011
=5.37 %
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