Question

1. The pKa of boric acid is 9.24. What is the value of the dissociation constant?

2. What molarity of botanic acid (pKa = 4.82) would have a pH of 3.0?

3. A 0.08 M solution of an acid is 0.04% ionized. What is the pKa?

4. Butyric acid has a pKa of 4.82.

a) what is the pH of 0.20 Butyric acid?

b) what is the value of pKb for this compound?

c) using the pKb from part b) to find pOH, determine the pH of 0.02 M sodium butanoate.

5. What is the concentration of the OH- ion in 0.05 M Hal?

6. What is the pH of the solution that results from mixing 8.0 mL of 0.30 M NaOH and 90mL of 0.10 M lactic acid? (pKa of lactic acid = 3.86)  

Answer 1

1.

p^Ka of boric acid = 9.24

K_a of boric acid = ?

Formula: -

ie.

thus,

  

--------------------------

2.

p^Ka of butanoic acid = 4.82

p^H of butanoic acid = 3

Molarity (concentration) of butanoic acid = Z = ?

First calculate calculate K_a

we kave,

thus,

ie.

since, butanoic acid is weak acid, it dissociate as fallow

...(A)

NOW, construct an ICE table for equilibrium (A)

initial	Z		0		0
change	- x		+ x		+ x
equilibrium	Z - x		x		x

The K_{a} expression for equilibrium (A) is

  

i.e.   ....(1)

we have, p^H = 3

we know that,

i.e.

thus,

i.e.

Now , from equation (1) we have

  

i.e.

i.e.

i.e.  

i.e.

therefore, molarity of butanoic acid is 0.067 M.

--------------------------------------------

3.

Consider the given weak acid is HA

Thus Molarity of HA = 0.08 M

% ionisation of HA is = 0.04%

p^Ka of HA = ?

The dissociation of HA is

   ....(B)

Now, an ICE table for equilibrium (B) is

	HA		H+	+	A-
initial	0.08		0		0
change	- x		+ x		+ x
equilibrium	0.08 - x		x		x

The K_a expression for equilibrium (B) is

i.e. ....(2)

Now, acid ionized (x) is calculated as

  

ie.

i.e.

i.e.

therefore,

from equation (2) we have,

i.e

i.e.

Now,

  

i.e.

i.e.

i.e.