# Answer to Question #175527 in General Chemistry for ary

Question #175527

1. A 25.00 mL 0.015 M acetic acid was titrated with 15.00 mL of Mg(OH)2 to reach equivalent point. Calculate

(Ka HCOOH = 1.85 x 10-5)

a. the concentration of Mg(OH)2?

b. The pH at equivalence point.

c. The pH after 20% and 75% titration.

d. The pH change after 5 mL, 0.01 M HCl is added into the resultant solution in part

(b).

e. The pH change after 2 mL, 0.02 M NaOH is added into the resultant solution in part

(b).

1
2021-04-01T02:30:25-0400

a) equation of reaction is

2CH3COOH + Mg(OH)2 --> Mg(CH3COO-)2 + 2H2O

Concentration of acid(Ca) = 0.015M, Volume of acid(Va)= 0.025L, Volume of base(Vb) = 0.015L, Concentration of base(Cb)=? , Mole of acid(Na) = 2mol, mole of base(Nb) = 1mol

Cb = CaxVaxNb/NaxVb

Cb= 0.015x0.025x1/2x0.015

Cb= 0.0125M

The concentration of Mg(OH)2 is 0.0125M.

b)amount of Mg(CH3COO-)2 = 3.75x10^-4mol

Concentration of Mg(CH3COO-)2 = 0.000375/0.04 = 0.009375M

Let the concentration of Mg(CH3COOH)2 be X and OH- be X, then the final concentration of Mg(CH3COO-)2 will be (0.009375-X)

Kw= KaKb

Kb= 1x10^-14/1.85x10^-5

Kb= 5.4x10^-10

X=2.25x10^-6 M

pOH= -log[2.25x10^-6] = 5.64

pH= 14-5.64= 8.35

pH at equivalence point is 8.35

C) at 0%, amount of H+ = 0.015Mx0.025L = 0.000375mol

Volume of OH- at 20% = 20x0.015/100 = 0.003L

Amount of OH- at 20% titration = 0.0125 x 0.003 = 0.0000375mol

Amount of acid left= 0.000375-0.0000375 = 0.0003375mol

Volume of titre = 0.025+0.003 = 0.028L

[H+]= 0.0003375/0.028 = 0.0121

pH = -log[H+] = 1.92

at 75% titration

Volume of OH- = 75x0.015/100= 0.01125L

Amount of OH-at 75% = 0.0125x0.01125= 0.0001406mol

Amount of acid left= 0.000375-0.0001406 = 0.0002344mol

Volume of titre = 0.025+0.01125= 0.03625L

[H+]= 0.0002344/0.03625= 0.006466

pH= -log[0.006466]

pH = 2.19

D) final concentration of H+ = 0

Final concentration of Mg(CH3COO-)2 = 0.0000437

Final concentration of Mg(CH3COOH)2 = 0.00005

pH= 8.35+ -log[0.00005/0.0000437]

pH= 8.29

e) final concentration of Mg (CH3COOH)2 = 0.000147

Final concentration of Mg (CH3COO-)2 = 0.00004

pH = 8.35+ -log[0.00004/0.000147]

pH= 8.92

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