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:
(3) pH = 5.36
Explanation:
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-]
∴ 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
