Experiment 9: Determination of Ka of a Weak Acid

Question 1

What weak acid did we use?

Answer 1

Vinegar, acetic, HC2H3O2

Question 2

What strong base did we use?

Answer 2

NaOH

Question 3

Will the equivalence point be above 7, below 7, or at 7?

Answer 3

Above 7, the salt formed will be basic (weak acid + strong base)

Question 4

The graph looks like____ where the equivalence point is…?

Answer 4

A sideways s. Equivalence point occurs when moles base= moles acid, when the slope is undefined. This for us is above 7

Question 5

In water acids dissociate like…?

Answer 5

HA + H2O –> A- + H3O.

Question 6

Ka=

Answer 6

products over reactants

Ka= [H3O]^2/[HA]

Question 7

To find the pH at the half equivalence point you must….

Answer 7

take the volume of base at equivalence, divide by 2, then find the pH at that volume.
pH1/2=pKa

Question 8

The conjugate acid is…

Answer 8

The product side molecule that gained the proton

Question 9

The conjugate base is….

Answer 9

The product side value that lost the proton

Question 10

Hendersen hasselback equation

Answer 10

pH=pKa+log[A^-/HA]

Question 11

If HX is the formula of a weak acid, what is the formula of it’s conjugate base?

Answer 11

X^-