Find the pH values for the points along the titration the titration of 25 mL of

Question

Find the pH values for the points along the titration the titration of 25 mL of a 0.1 M solution of maleic acid (HO_2CCHCHCHCO_2H) titrated with 0.10 M solution of NaOH. The pK_a values of maleic acid are pK_a1 = 1.92. pK_a2 = 6.27. What chemical reaction is happening at 0.00 mL added? What chemical reaction is happening at 1.00 mL added? What chemical reaction is happening at 25.00 mL added? What chemical reaction is happening at 26.00 mL added? What chemical reaction is happening at 50.00 mL added? What chemical reaction is happening at 52.00 mL, added?

Explanation / Answer

a)

when 0 ml added... this is a weak acid:

H2A = H+ + HA- Ka1 mainly

and

HA- = H+ A-2 Ka2 very low in concentrations

after 1 mL of base added...

NaOH + H2A = HA- + H2O

there is

H2A >> HA- this is a buffer

pH = pKa + log(HA-/H2A)

c)

after V = 25 mL is added

mmol of base = MV = 0.1*25 = 2.5 mmol of OH-

mmol of acid = 25*0.1 = 2.5 mmol of H2A

this is FIRST equivalence point

so

NaOH + H2A = HA- + H2O

there is ONLY HA- in solution

so

HA- + H2O = H2A

pH = 1/2*(pKa1+pKa2)

d)

after 26 mL

there is no more H2A left

only

HA- >> and A-2

this is a buffer, second ionization region

pH = pKa2+ log(A-2/HA-)

finally

after 50 mL is added

this is complete neutralization

HA- + NAOH = A-2 + H2O

so

A-2 + H2O hydrolyses

A-2 + H2O = HA- + OH-

after 52 mL

this is mainly OH- in excess, so

OH- will give us most of pOH

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.