a student following the procedure in this experiment prepared three solutions by

Question

a student following the procedure in this experiment prepared three solutions by adding 11.50, 23.00, and 34.50 mL of a 7.60 x 10-2 M NaOH sloution to 50.00 mL of a 0.112M soution of a weak acid. The solutions were labeled 1,2,and 3, respectively. Each of the solutions was diluted to a total volume of 100.00 mL with distilled water. The pH readings of these solutions were: 1)6.42, 2) 6.81, and 3) 7.09

1) convert the pH of each solution to an equivalent H3O+ ion concentration.

2) For each solution, Caluate the number of moles of acid added.

3) For each solution, caluate the number of moles of OH- ion added.

4) For each solution, determine the number of moles of HAn and of An- ion at equilibrium.

5) For each solution, caluate the HAn and An- ion molar equilibrium concentrations.

6) Determine the reciprocal of the An- ion concentration for each solution.

Explanation / Answer

1)pH=-log[H3O+]

[H3O+]=10^-pH

for 6.42[H3O+]=10^-6.42

for 6.81 [H3O+]=10^-6.81

for 7.09 [H3O+]=10^-7.09

2)for each solution same no of moles of acid are added=50*0.112 millimoles=5.6 millimoles

3)no of moles of OH- added=11.5*7.6*10^-2,23*7.6*10^-2,34.5*10^-2=0.874 millimoles,1.748 millimoles,2.622 millimoles

4) and 5)concentration moles /volume

volume=100 ml

for solution 1

moles of An-=moles of base=0.874 millimoles

moles of HAn=5.6-0.874=4.726 millimoles

concentration An=0.874/100=0.000874

concentration HAn=4.726/100=0.04726

for solution 2

moles of An-=moles of base=1.748 millimoles

moles of HAn=5.6-1.748=3.852 millimoles

concentration An=1.748/100=0.001748

concentration HAn=3.852/100=0.03852

for solution 3

moles of An-=moles of base=2.622 millimoles

moles of HAn=5.6-0.874=2.978 millimoles

concentration An=2.622/100=0.002622

concentration HAn=3.852/100=0.038522

6)

reciprocal=1/0.000874,1/0.001748,1/0.002622=1144.16,572.08,381.38

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