Question

Calculate the pH at the equivalence point in titrating 0.071 M solutions of each of the following with 0.028 M NaOH.

(a) hydroiodic acid (HI) pH =___

(b) hypochlorous acid (HClO), Ka = 3e-08 pH =____

(c) hydrosulfuric acid (H2S), Ka = 9.5e-08 pH =____

Answer 1

a) Since the you are addind a strong base into a strong acid the pHa te the equivalence point is 7

pH=7

b)The pH at the equivalence point is when the acid has been neutralized and almost 100% of the conjugate base is in solution, so you can assume at this point the pH is given by the conjugate base, you sould alos take into account that during the titration you are going to dilute your original concentration, a rough estimate of this dilution is:

since you need to add the same amount (in moles) of base in order to neutralize the acid, you need to add a volume of base 2.53 times bigger than the original value, using the following formula you can calculate the concentration at the equivalent point:

----->C2 and V2 are at the equivalence ponit, V2=V1+2.53V1 (since you are adding a volume of base 2.53 bigger than the acid)

based on the following reaction and ICE table you can calculate the pH:

Initial	0.020			0	0
Change	-X			+X	+X
Equilibrium	0.020-X			X	X

Using the mass action equation for Kb:

and using the concentrations at equilibrium:

<<- this yields a second degree equation:

and solving with the quadratic formula:

<<<-.. this is the concentration of OH

c) H2S is a diprotic acid, since there is only 1 Ka I am assuming you just need the first equivalence point, and also the second Ka for this acid is near 14 so its hard to calculate the second equivalence point.

The procedure is just as above, the concentration at the equivalence point is the same (0.020),

based on the following reaction and ICE table you can calculate the pH:

Initial	0.020			0	0
Change	-X			+X	+X
Equilibrium	0.020-X			X	X

Using the mass action equation for Kb:

and using the concentrations at equilibrium:

<<- this yields a second degree equation:

and solving with the quadratic formula:

<<<-.. this is the concentration of OH