Question

Consider two separate titrations:

Titration 1: 50 mL of 0.1 M acetic acid ($CH_3COOH$) is titrated with 0.1 M NaOH. Titration 2: 50 mL of 0.1 M HCl is titrated with 0.1 M $NH_3$.

Given the following information:

• For $CH_3COOH$, $K_a = 1.8 \times 10^{-5}$ ($pK_a \approx 4.74$).
• For $NH_3$, $K_b = 1.8 \times 10^{-5}$ ($pK_b \approx 4.74$, $pK_a(NH_4^+) \approx 9.26$).
• Ionic product of water, $K_w = 1.0 \times 10^{-14}$.
• Indicator pH ranges (approximate visible color change):
  • Methyl Orange (MO): pH 3.1 - 4.4 (Red to Yellow)
  • Methyl Red (MR): pH 4.2 - 6.3 (Red to Yellow)
  • Phenolphthalein (PhPh): pH 8.2 - 10.0 (Colorless to Pink)
• Assume the color change of an indicator is perceived when the ratio of the indicator's ionized form to unionized form is between 0.1 and 10. The pH range for color change is approximately $pK_{In} \pm 1$. Use $pK_{In}$ values: MO = 3.7, MR = 5.0, PhPh = 9.6.

Identify the correct statement(s):

• A. Phenolphthalein is a suitable indicator for Titration 1.
• B. Methyl Red is a suitable indicator for Titration 2.
• C. In Titration 1, after adding 25 mL of NaOH, a solution containing Methyl Orange will appear Yellow.
• D. In Titration 2, after adding 49.9 mL of $NH_3$, a solution containing Methyl Red will appear Red.

Answer 1

Answer: (A), (B), (C), (D)

(A), (B), (C), (D)

Answer 2

Let's analyze each titration and option:

Titration 1: 50 mL 0.1 M $CH_3COOH$ vs 0.1 M NaOH (Weak Acid vs Strong Base)

• Equivalence Point: Moles of $CH_3COOH = 50 \times 0.1 = 5$ mmol. Moles of NaOH needed = 5 mmol. Volume of NaOH needed = 5 mmol / 0.1 M = 50 mL. Total volume = 100 mL. At equivalence, the solution contains 5 mmol of $CH_3COONa$ in 100 mL, so $[CH_3COONa] = 0.05$ M. $CH_3COONa$ is a salt of a weak acid and strong base, so the solution will be basic due to hydrolysis of $CH_3COO^-$. $CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$ $K_h = K_w / K_a = 10^{-14} / (1.8 \times 10^{-5}) \approx 5.56 \times 10^{-10}$. $[OH^-] = \sqrt{C \times K_h} = \sqrt{0.05 \times 5.56 \times 10^{-10}} = \sqrt{2.78 \times 10^{-11}} \approx 5.27 \times 10^{-6}$ M. $pOH = -\log(5.27 \times 10^{-6}) \approx 5.28$. $pH = 14 - pOH = 14 - 5.28 = 8.72$. The equivalence point for Titration 1 is at pH $\approx 8.72$.

• Option A: Suitability of Phenolphthalein for Titration 1. The pH range of PhPh is 8.2 - 10.0. The equivalence point pH (8.72) falls within this range. The steep part of the titration curve for a weak acid-strong base titration occurs around the equivalence point pH. Since the equivalence point pH falls within the indicator's range, Phenolphthalein is a suitable indicator for Titration 1. Therefore, option (A) is correct.

• Option C: Color of Methyl Orange after adding 25 mL NaOH in Titration 1. 25 mL NaOH is half the equivalence volume (50 mL). At the half-equivalence point, $[CH_3COOH] = [CH_3COO^-]$. This forms a buffer solution. $pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]} = pK_a + \log(1) = pK_a = 4.74$. The pH of the solution after adding 25 mL NaOH is 4.74. The pH range for Methyl Orange color change is 3.1 - 4.4 (Red to Yellow). At pH 4.74, which is above the upper limit of the range (4.4), the indicator will be predominantly in its basic (ionized) form, which is Yellow. Therefore, option (C) is correct.

Titration 2: 50 mL 0.1 M HCl vs 0.1 M $NH_3$ (Strong Acid vs Weak Base)

• Equivalence Point: Moles of HCl = 50 \times 0.1 = 5 mmol. Moles of $NH_3$ needed = 5 mmol. Volume of $NH_3$ needed = 5 mmol / 0.1 M = 50 mL. Total volume = 100 mL. At equivalence, the solution contains 5 mmol of $NH_4Cl$ in 100 mL, so $[NH_4Cl] = 0.05$ M. $NH_4Cl$ is a salt of a strong acid and weak base, so the solution will be acidic due to hydrolysis of $NH_4^+$. $NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+$ $K_h = K_w / K_b = 10^{-14} / (1.8 \times 10^{-5}) \approx 5.56 \times 10^{-10}$. $[H_3O^+] = \sqrt{C \times K_h} = \sqrt{0.05 \times 5.56 \times 10^{-10}} = \sqrt{2.78 \times 10^{-11}} \approx 5.27 \times 10^{-6}$ M. $pH = -\log(5.27 \times 10^{-6}) \approx 5.28$. The equivalence point for Titration 2 is at pH $\approx 5.28$.

• Option B: Suitability of Methyl Red for Titration 2. The pH range of Methyl Red is 4.2 - 6.3. The equivalence point pH (5.28) falls within this range. The steep part of the titration curve for a strong acid-weak base titration occurs around the equivalence point pH. Since the equivalence point pH falls within the indicator's range, Methyl Red is a suitable indicator for Titration 2. Therefore, option (B) is correct.

• Option D: Color of Methyl Red after adding 49.9 mL $NH_3$ in Titration 2. Moles of HCl = 5 mmol. Moles of $NH_3$ added = 49.9 mL $\times$ 0.1 M = 4.99 mmol. HCl is the limiting reactant initially, but after adding $NH_3$, $NH_3$ reacts with HCl. We have excess HCl before the equivalence point. Moles of HCl remaining = Moles initial HCl - Moles $NH_3$ added = 5 mmol - 4.99 mmol = 0.01 mmol. Total volume = 50 mL (HCl) + 49.9 mL ($NH_3$) = 99.9 mL. $[H^+] = \frac{\text{Moles HCl remaining}}{\text{Total volume}} = \frac{0.01 \text{ mmol}}{99.9 \text{ mL}} \approx 1.001 \times 10^{-4}$ M. $pH = -\log(1.001 \times 10^{-4}) \approx 4.0$. The pH of the solution after adding 49.9 mL $NH_3$ is approximately 4.0. The pH range for Methyl Red color change is 4.2 - 6.3 (Red to Yellow). At pH 4.0, which is below the lower limit of the range (4.2), the indicator will be predominantly in its acidic (unionized) form, which is Red. Therefore, option (D) is correct.

All four options (A), (B), (C), and (D) are correct.

Explanation: Option (A) is correct because the equivalence point pH of the weak acid-strong base titration (pH 8.72) falls within the color change range of Phenolphthalein (pH 8.2-10.0). Option (B) is correct because the equivalence point pH of the strong acid-weak base titration (pH 5.28) falls within the color change range of Methyl Red (pH 4.2-6.3). Option (C) is correct because at 25 mL NaOH addition in Titration 1, the pH is equal to the $pK_a$ of acetic acid (4.74). This pH is above the color change range of Methyl Orange (3.1-4.4), so the indicator is in its basic (Yellow) form. Option (D) is correct because at 49.9 mL $NH_3$ addition in Titration 2, the solution is slightly before the equivalence point and contains excess HCl, resulting in an acidic pH (approx. 4.0). This pH is below the color change range of Methyl Red (4.2-6.3), so the indicator is in its acidic (Red) form.