(b) the conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 s m–1. if the molar conductance at infinite dilution (λ°) of aqueous sodium chloride, sodium formate and hydrochloric acid are 1.264 × 10–2, 1.046 × 10–2 and 4.261 × 10–2 respectively at 25°c determine the acid dissociation constant and the pka for the acid.

Explanation:

The given data is as follows.

   (NaCl) =

   (H-O=C-ONO) =

   (HCl) =

Conductivity of monobasic acid is

     Concentration = 0.01

Therefore, molar conductivity () of monobasic acid is calculated as follows.

                                  =

                                 =

                                 =

Also, =

                            =

                            =

Relation between degree of dissociation and molar conductivity is as follows.

                             =

                             = 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

                     K =

Putting the values into the above formula we get the following.

                     K =

                        =

                        =

                       =

Hence, the acid dissociation constant is .

Also, relation between and is as follows.

                              =

                              = 3.7454

Therefore, value of is 3.7454.

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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