Question

Calculate the pH in the titration of 50mL of a .2M acetic acid and .312M NaOH Ka = 1.8x10^-5

- 5 mL before equivalence point

Answer 1

CH₃COOH + NaOH ---> CH₃COONa +H₂O

From the balanced equation, we get that 1 mol of acid is needed to react completely with 1 mole of the base.

Hence, M_AV_A = M_BV_B

(0.2 M)(50 mL) = (0.312 M) V_B

solving the above equation, we get:

V_B = 32.1 mL

Hence, 32.1 mL of base has to be added to reach equivalence point.

5 mL before equivalence point = 32.1 - 5 = 27.1 mL

When 27.1 mL of NaOH is added, amount of base reacted is:

(0.312 mol/L)(0.0271 L) = 0.00846 mol

amount of acid reacted = amount of base reacted = 0.00846 mol

amount of acetate ion formed = 0.00846 mol

volume of solution = 50 + 27.1 = 77.1 mL

[CH₃COO^-] = 0.00846 mol/77.1 mL

amount of acid left unreacted =(0.2 mol/L)(0.050 L) - 0.00846 mol =0.00154 mol

CH₃COOH -----> CH₃COO^- + H⁺

Let x moles of acetic acid dissociate.

At equilibrium:

[CH₃COOH] = (0.00154 mol - x)/0.050 L 0.00154 mol/0.0771 L

[H⁺] = x/0.0771 M

[CH₃COO^-] = (x/0.0771) + (0.00846/0.0771)M = (x+0.00846)/0.771

K_a = 1.8 x 10⁵

x² + 0.00846x = 2.137 x 10^-9

x = 2.51 x 10^-7 mol

[H⁺] = (2.51 x 10^-7 mol)/0.0771 L = 3.25 x 10^-6 M

pH = -log[H⁺] = -log(3.25 x 10^-6) = 5.49

Therefore, at 5mL before equivalence point, pH is 5.49.

Therefore,

[H⁺] = x/0.050 M = 3.72 /0.050 =