2. Ka of a Weak Acid

Question 1

Weak acids react with water to create this equilibrium

Answer 1

HA +H2O H3O+ + A-

Question 2

The acid dissociation constant represents:

Answer 2

The equilibrium of the weak acid with water

Question 3

pH=

Answer 3

-log[H+]

Question 4

pKa=

Answer 4

-log(Ka)

Question 5

Titration Curve:

Weak acid-strong base:

Answer 5

buffer region___//||—, indicator is phenolphthalein (ph 8.3-10)

Question 6

Titration Curve:

Strong acid-weak base

Answer 6

Indicator: methyl orange (pH=3.1-4.4)

Question 7

Titration Curve:

weak acid-weak base

Answer 7

NAH YA BISH

Question 8

After each addition of base, equilibrium is reestablished according to:

Answer 8

Ka=[H3O+]*[A-]/[HA]

Question 9

pH=pKa+(…)

Answer 9

pH=pKa+ log([A-]/[HA])

[A-]=[HA]—> pH=pKa

Question 10

Method to determine Ka (1/2)

Answer 10

Measure the pH of something with a known weak acid concentration

Question 11

Method to determine Ka (2/2)

Answer 11

Measure the pH at the half-neutralization point in the titration of a weak acid with a strong base

Question 12

At the beginning of the titration, what species are present?

Answer 12

HA and a VERY small amount of H3O and A- from its ionization

Question 13

When does pH=pKa?

Answer 13

At the half-equivalence point….(half way before dramatic rise)

pH1/2=pKa

Question 14

Equilibrium is reestablished after each addition of base. True/false?

Answer 14

True. you could therefore get a Ka value for any point before the equivalence point

Question 15

The first point on a titration curve represents….

Answer 15

The pH of the Acid

Question 16

At the equivalence point…

Answer 16

The moles of acid = moles of base

Question 17

In the titration of 10.00mL of an unknown weak acid with a strong base that has a molarity of 0.1002M, the equivalence volume was found to be 19.20mL.

Answer 17

Calculate the concentration of unknown weak acid:
mole acid=mole base
(10.00mL)(x) = (0.1002M)(19.20 – 10.00)
x = 0.1923M

If the initial pH of the solution was recorded as 3.12, what is the Ka and pKa of the unknown weak acid:
-log(3.12) = 7.59x10-4
= Ka = 2.99 x 10-6
-log(2.99 x 10-6) = pKa = 5.52