25 mL of 0.1 M lactic acid (CH3CH(OH)COOH) is titrated with 0.1 M sodium hydroxi

Question

                    25 mL of 0.1 M lactic acid (CH3CH(OH)COOH) is titrated with 0.1 M sodium hydroxide (NaOH). The titration curve is shown.                 

                    
                

                    (a) What is the approximate pKa?                 

                    (b) At the pKa, what fraction of the carboxyl group will have been converted to COO-?                 

25 mL of 0.1 M lactic acid (CH3CH(OH)COOH) is titrated with 0.1 M sodium hydroxide (NaOH). The titration curve is shown. What is the approximate pKa? At the pKa, what fraction of the carboxyl group will have been converted to COO-?

Explanation / Answer

a) ph = pka + log [sodium lactate]/[lactic acid]

Lactic acid + NaOH <-----------> sodium lactate ,

when NaOH moles = 1/2 moles of acid then [sodium lactate] =[lactic acid] and hence pH = pka ,

hence vol of NaOH = 12.5 ml and pH = pka = 3.2 (nearly)

b) at this pka half Carboxyl group will be converted to COO_

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

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