Titration of Weak Acid Workshop Sheet Stuff

The following is adapted from problem #10-54 in Oxtoby, Gillis, & Nachtrieb.

A sample of 50.00 mL of 0.1281 M aqueous solution of chloroacetic acid, CH₂ClCOOH (K_a = 1.4 ´ 10^-3) is titrated with a 0.08971 M aqueous NaOH.

a) calculate the equivalent volume, V_eq.
 

The equivalent volume is the volume at which enough titrant has been added to completely react with the analyte. In this case, the reaction is a simple neutralization:

So, at the equivalent volume, the number of moles of OH^- added will be equal to the number of moles of HA present initially. Therefore:

This states that when 71.40 mL of NaOH solution are added to the acid solution, the acid is completely neutralized, but no excess base has been added.

b) calculate the pH before any NaOH is added.
 

Since no base has been added, this is a simple weak acid problem. The reaction of the acid with the water is:

This is solved as any other equilibrium problem, that is, with a table of equilibrium concentrations.
 
 
 

	HA	H3O+	A-
initial	0.1281	0	0
change	-x	x	x
equilibrium	0.1281-x	x	x

x = 1.339´ 10^-2.

This is a bit high for the assumption that 0.1281 - x » 0.1281. We can either solve the initial problem using the quadratic equation or iterate. Since you know how to use the quadratic, let's iterate. Use the value of x you just estimated as a better approximation of the final concentration of the acid in the K_a expression as follows. Using a value of x = 1.339 ´ 10^-3, we can write:

This is an "improved" estimate of the equilibrium concentration of A^- and H₃O⁺. To insure that we have converged on a solution, a third iteration can be performed, thereby improving the estimate even more.

 
 

x = 1.271 ´ 10^-2

The last two estimates are equivalent to three significant figures, so no more iterations are required.
Since x = [H₃O⁺], pH = -log (1.27 ´ 10^-2) = 1.90

c) calculate the pH at 0.5 V_eq.
 
 

As the neutralization progresses, acid is being converted to conjugate base by the reaction
 
 

Since the solution contains both the acid and the conjugate base, we can use the Henderson-Hasselbach equation. To keep track of the relative levels of the acid and conjugate base, it helpful to set up a table of moles (not concentrations as is done when determining equilibrium concentrations; the use of moles instead of concentration in the Henderson-Hasselbach equation is not necessary but is more convenient). The number of moles of hydroxide is obtained by multiplying the volume, 0.5Veq (35.70 mL), by the concentration, 0.08971 M.
 
 
 

Using these values in the Henderson Hasselbach equation yields,

d) calculate the pH at 0.9 V_eq.

This part is solved in exactly the same manner as part c. The table is:

e) calculate the pH at 0.99 V_eq.
 

This part is solved in exactly the same manner as parts c and d. The table is:
 
 

moles	HA	OH-	A-
initial	6.405 x 10-3	6.341 x 10-3	0
change	-6.341 x 10-3	-6.341 x 10-3	6.341 x 10-3
equilibrium	0.064 x 10-3	0	6.341 x 10-3

f) calculate the pH at V_eq.

At the equivalence point, all of the original acid has been converted into its conjugate base. To find the pH at this point, the reaction that needs to be considered is that between the conjugate base and water. This is,

The equilibrium concentrations of these species are found as in any typical equilibrium problem. The initial concentration of the base, however, is not the same as the initial concentration of the weak acid. The number of moles of the conjugate base is the same as the initial number of moles of the weak acid (there is a 1:1 stoichiometric relationship), but the volume is greater because at the equilvalence point, 71.4 mL of base solution have been added to the original volume. Thus the total volume is 121.40 mL, yielding a concentration of 0.05276 M.

The table of concentrations is:

 
In this case, x << 0.05276, so our assumption that it is very small is good.
 

To find the pH,

 

pH = -log [H₃O⁺] = 7.79

g) calculate the pH at 1.01 V_eq.
 

Beyond the endpoint, the pH will be determined by the amount of excess hydroxide that is added. To find the concentration, subtract the number of moles of base that was consumed by the neutralization from the total number of moles added. Using the same type of table used in part c, we can write,
 
 

moles	HA	OH-	A-
initial	6.405 x 10-3	6.469 x 10-3	0
change	-6.405 x 10-3	-6.405 x 10-3	6.405 x 10-3
equilibrium	0	0.0064 x 10-3	6.405 x 10-3

The hydroxide concentration is simply the moles of hydroxide divided by the total volume (122.11 mL). So,

 

 

pH = -log [H₃O⁺] = 10.71

h) calculate the pH at 1.5 V_eq.
 
 
 

The hydroxide concentration is simply the moles of hydroxide divided by the total volume (122.11 mL). So,

pH = -log [H₃O⁺] = 12.31

i) make a rough sketch of the titration curve (pH vs. volume NaOH added) based on your calculated pH values.