Question

USE THE FOLLOWING SCENARIO FOR QUESTIONS 6-10. A 25-ml sample of 0.10 Macetic acid is titrated with 15-ml 0.10 M NaOH. (K, for acetic acid - 1.8 x 10-) Titration curve Vol.Titrant added 6. Which of the following balanced equations represents the equilibrium at the 15-ml point in the titration? A. CH3COOH(aq) + H2O(1) = H30*(aq) + CH,COO"(aq) B. CH3COOH(aq) + NaOH(aq) + H2O(l) + NaCH3COO(aq) C. CHC00 (aq) + H20(I) = CH3COOH(aq) + OH"(aq) D. NaOH(aq) + H2O(0) = Na (aq) + OH(aq) + H2O(1) 7. What is the pH after 15 mL of NaOH have been added? A. 3.79 B.4.92 C. 6.34 D. 11.61 8. At which point on the curve provided would the pH = 4.74? A. A В. В C.C D. You can't determine this without knowing the volume of base added. 9. Which of the following balanced equations represents the equilibrium at the equivalence point in the titration? A. CHCOOH(aq) + H2O(1) = H30*(aq) + CH,COO"(aq) B. CHCOOH(aq) + NaOH(aq) = H2O(l) + NaCH3COO(aq) C. CH,COO"(aq) + H20(I) = CH3COOH(aq) + OH"(aq) D. NaOH(aq) + H2O(l) = Na"(aq) + OH(aq) + H2O(U) 10. Which of the following COULD BE the pH at the equivalence point of this titration (this shouldn't require calculations)? A. 1.0 B. 2.87 C. 8.72 D. 13.0
Can anyone check my answers, and correct me if I’m wrong?

Answer 1

6.

At equivalence point

mmoles of CH₃COOH =  mmoles of NaOH

At 15 mL NaOH , the solution contains excess acetic acid.

hence equilbrium is due to dissociation of acetic acid.

CH₃COOH(aq) + H₂O(l)  $\rightleftharpoons$ H₃O⁺ (aq) + CH₃COO^- (aq)

Option A is correct.

7.

When moles of 15 ml 0.1 M NaOH is added

mmoles of NaOH = volume* molarity = 15*0.1= 1.5

mmoles acetic acid (AcOH) = (25*0.1) = 2.5

Excess mmoles of acid = (2.5 - 1.5) = 1

mmoles of sodium acetate (NaOAc) formed = moles of NaOH added = 1.5

Ka = 1.8*10^-5, pKa = -log Ka = - log(1.8*10^-5) = 4.74

Now the solution is an acid buffer

Using Hendersson -Hasselbalch equation.

pH = pKa + log [NaOAc/AcOH]

= 4.74 + log (1.5/1)

= 4.74 + 0.18 = 4.92 .

Correct answer is B.

8.

At half equivalence point

[NaOAc] = [AcOH]

pH = pKa

The point A is the half equivalence point.

9.

At the equivalence point , pH is due to dissociation of sodium acetate.

Hence equilibrium reaction is

CH₃COO^- (aq)  + H₂O (l) $\rightleftharpoons$ CH₃COOH (aq) + OH^- (aq).

10.

At equivalence point , pH of the solution is due to dissocistion of the salt (sodium acetate).

For salt of weak acid strong base , pH at equivalence point

pH = 7 + $\frac{1}{2}$ (pka + logC)

at , equivalence point mmoles of salt = 25*0.1= 2.5

Volune of solution = (25+25)= 50.

Then , concentration of salt (c) = (2.5/50)= 0.05

Now,pH = 7 + $\frac{1}{2}$ [4.74 + log(0.05)]

Or, pH = 7 + 1.72 = 8.72 .

Option C is correct.