Question

Find the pH during the titration of 20.00 mL of 0.2290 M benzoic acid, C₆H₅COOH (K_a = 6.3 ✕ 10^-5), with 0.2290 M NaOH solution after the following additions of titrant.

(a)    0 mL =


(b)    10.00 mL =


(c)    15.00 mL =


(d)    19.00 mL =


(e)    19.95 mL =


(f)    20.00 mL =


(g)    20.05 mL =


(h)    25.00 mL =

Please help..!!!

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Answer 1

Let HA be the shorthand for butanoic acid
A)
At 0 ml NaOH:
0.2290 M HA
Ka = 6.3x10^-5 = [H+][A-]/[HA]
let X = [H+]
6.3x10^-5 = [H+][A-]/[HA] = X^2 / (0.2290 - X)
assume X << 0.2290 M
X = 0.003798 = [H+]

pH = - log [H+]
pH = 2.42

At 10.0, 15.00, 19.00, and 19.95 mL NaOH added you have a buffer solution composed of HA and A^-1. Use the Henderson-Hasselbalch equation-
pH = pKa + log(A-/HA)
B)
at 10.00 mL NaOH added HA=A- so pH = pKa
pH = -log(6.3x10^-5) = 4.2
C)
at 15.00 mL NaOH added, 3/4 of HA has been converted to A-
pH = pKa + log(0.75/0.25) = 5.29
D)
at 19.00 mL NaOH added
started with 4.58 millimoles of HA (20.00 mL x 0.229 M = 4.58 mmol)
added 4.35 millimol NaOH (19.00 mL x 0.2290 M = 4.35)
now have 4.35 millimol A- and 0.229 millimol HA
pH = pKa + log(4.35/0.229) = 5.84