Question

In an acid-base titration experiment, 50 mL of a 0.05 M solution of acetic acid (Ka= 1.75 x 10-5 ) was titrated with a 0.05M solution of NaOH at 25 ºC. The system will acquire this pH after addition of 40 mL of the titrant:please show work.(1) 3.27; (2) 4.58 (3) 5.36 (4) None of the above

Answer 1

Answer:

(3) 5.36

Explanation:

Since this is a titration of a weak acid before reaching equivalence point, we will have effectively a buffer solution. Then we can use the Henderson-Hasselbalch equation to answer this question.

The reaction is:

HAc + NaOH ⇒ NaAc + H₂O

V NaOH = 40 mL x 1 L/1000 mL = 0.040 L

mol NaOH reacted with HAc = 0.040 L x 0.05 mol/L = 0.002 mol

mol HAC originally present = 0.050 L x 0.05 mol/L = 0.0025 mol

mol HAc left after reaction = 0.0025 - 0.002 = 0.0005

Now that we have calculated the quantities of the weak acid and its conjugate base in the buffer, we just plug the values into the equation

pH = pKa + log ((Ac⁻)/(HAc))

(Notice we do not have to calculate the molarities of  Ac⁻ and HAc because the volumes cancel in the quotient)

pH = -log (1.75 x 10⁻⁵) + log (0.002/0.0005) = 5.36

THe answer is 5.36

Answer 2

Answer:

(3) pH = 5.36

Explanation:

• CH3COOH + NaOH ↔ CH3COONa + H2O

after 40 mL NaOH:

∴ C CH3COOH = ((0.050 L)(0.05 mol/L) - (0.040L)(0.05 mol/L)) / (0.090 L)

⇒ C CH3COOH = 5.55 E-3 M

∴ C CH3COONa = ((0.040 L)(0.05 M)) / (0.090 L) = 0.022 M

mass balance:

⇒ 0.022 + 5.55 E-3 = [CH3COOH] + [CH3COO-] = 0.02755 M

charge balance.

⇒ [H3O+] + [Na+] = [CH3COO-] + [OH-]......[OH-] its come from water

⇒ [H3O+] + 0.022 M = [CH3COO-]

• CH3COOH + H2O ↔ CH3COO-  +  H3O+

∴ Ka = 1.75 E-5 = [H3O+] [CH3COO-] / [CH3COOH]

∴ [CH3COOH] = 0.02755 - [CH3COO-]

⇒ [CH3COOH] = 0.02755 - ( [H3O+] + 0.022 )

⇒ [CH3COOH] = 5.55 E-3 - [H3O+]

⇒ Ka = [H3O+] ( 0.022 + [H3O+] ) / (5.55 E-3 - [H3O+] ) = 1.75 E-5

⇒ [H3O+]² +0.022[H3O+] = 9.7125 E-8 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.02202[H3O+] - 9.7125 E-8 = 0

⇒ [H3O+] = 4.414 E-6 M

∴ pH = - Log [H3O+]

⇒ pH = 5.3552 ≅ 5.36