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 10-2M NaOH solution to 50.00 mL of a .112M solution 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 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, calculate the number of moles of acid added.
3. For each solution, calculate the number of moles of OH- ion added.
4. For each solution, determine the number of moles of HAn and An- ion at equilibrium.
5. For each solution, calculate 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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