Question

1. methanoic acid is also called formic acid. it has the chemical formula hcooh(l). it is a colourless fuming liquid that is mainly used as a preservative. it exhibits the following equilibrium in water: hcooh(aq) + h₂o(l) → hcoo⁻(aq) + h₃o⁺(aq) if the ionization constant for the above acid is 1.8 x 10⁻⁴, the ph of methanoic acid in a 0.35 mol/l solution will be _______. (2 marks) 2) a 4.50 g tablet of magnesium hydroxide neutralizes 400.0 ml of stomach acid, hcl. what is the concentration of hcl in the stomach? (2 marks) 3) a 35.0 ml sample of (monoprotic) lactic acid, c₃h₆o₃, is titrated with 20.0 ml of a 4.0 x 10⁻⁴ mol/l sodium hydroxide solution. what is the ph of the resulting solution at the equivalence point, if kₐ for lactic acid is 1.4 x 10⁻⁴? (9 marks) 4) the concentration of the hydroxide ion, oh⁻(aq), in an aqueous solution is 2.5 x 10⁻³ mol/l. what is the ph of the solution? (2 marks)

Explanation:

Problem 1:

Step1: Define \( K_a \) for weak acid

For methanoic acid (\( \text{HCOOH} \)), a weak acid, the ionization is \( \text{HCOOH(aq)} + \text{H}_2\text{O(l)}
ightleftharpoons \text{HCOO}^-(\text{aq}) + \text{H}_3\text{O}^+(\text{aq}) \). The acid dissociation constant \( K_a = \frac{[\text{HCOO}^-][\text{H}_3\text{O}^+]}{[\text{HCOOH}]} \). Let \( x = [\text{H}_3\text{O}^+] = [\text{HCOO}^-] \) (since 1:1 ionization), and \( [\text{HCOOH}] = 0.35 - x \). Assume \( x \ll 0.35 \) (valid for weak acids with small \( K_a \)), so \( [\text{HCOOH}] \approx 0.35 \).

Step2: Solve for \( x \)

\( K_a = 1.8 \times 10^{-4} = \frac{x \cdot x}{0.35} \) → \( x^2 = 1.8 \times 10^{-4} \times 0.35 \) → \( x^2 = 6.3 \times 10^{-5} \) → \( x = \sqrt{6.3 \times 10^{-5}} \approx 7.94 \times 10^{-3} \, \text{mol/L} \).

Step3: Calculate pH

\( \text{pH} = -\log(x) = -\log(7.94 \times 10^{-3}) \approx 2.10 \).

Step1: Write neutralization reaction

\( \text{Mg(OH)}_2 + 2\text{HCl}
ightarrow \text{MgCl}_2 + 2\text{H}_2\text{O} \). Molar mass of \( \text{Mg(OH)}_2 \) is \( 24.31 + 2(16.00 + 1.01) = 58.33 \, \text{g/mol} \). Moles of \( \text{Mg(OH)}_2 = \frac{4.50 \, \text{g}}{58.33 \, \text{g/mol}} \approx 0.07715 \, \text{mol} \).

Step2: Relate moles of HCl

From reaction, 1 mol \( \text{Mg(OH)}_2 \) neutralizes 2 mol HCl. So moles of HCl \( = 2 \times 0.07715 = 0.1543 \, \text{mol} \). Volume of HCl is \( 0.400 \, \text{L} \).

Step3: Calculate concentration

Molarity \( M = \frac{\text{moles}}{\text{volume (L)}} = \frac{0.1543 \, \text{mol}}{0.400 \, \text{L}} \approx 0.386 \, \text{mol/L} \).

Step1: Moles of NaOH at equivalence

Moles of \( \text{NaOH} = 0.0200 \, \text{L} \times 4.0 \times 10^{-4} \, \text{mol/L} = 8.0 \times 10^{-6} \, \text{mol} \). At equivalence, moles of lactic acid (\( \text{HLac} \)) = moles of NaOH = \( 8.0 \times 10^{-6} \, \text{mol} \). Moles of \( \text{Lac}^- \) (conjugate base) = \( 8.0 \times 10^{-6} \, \text{mol} \). Total volume \( = 35.0 + 20.0 = 55.0 \, \text{mL} = 0.0550 \, \text{L} \).

Step2: Concentration of \( \text{Lac}^- \)

\( [\text{Lac}^-] = \frac{8.0 \times 10^{-6} \, \text{mol}}{0.0550 \, \text{L}} \approx 1.45 \times 10^{-4} \, \text{mol/L} \).

Step3: Hydrolysis of \( \text{Lac}^- \)

\( \text{Lac}^- + \text{H}_2\text{O}
ightleftharpoons \text{HLac} + \text{OH}^- \). Let \( K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.4 \times 10^{-4}} \approx 7.14 \times 10^{-11} \). Let \( x = [\text{OH}^-] = [\text{HLac}] \), \( [\text{Lac}^-] \approx 1.45 \times 10^{-4} - x \approx 1.45 \times 10^{-4} \) (since \( K_b \) is very small).

Step4: Solve for \( x \)

\( K_b = \frac{x \cdot x}{1.45 \times 10^{-4}} \) → \( x^2 = 7.14 \times 10^{-11} \times 1.45 \times 10^{-4} \) → \( x^2 = 1.035 \times 10^{-14} \) → \( x = \sqrt{1.035 \times 10^{-14}} \approx 1.017 \times 10^{-7} \, \text{mol/L} \).

Step5: Calculate pOH and pH

\( \text{pOH} = -\log(1.017 \times 10^{-7}) \approx 6.99 \), \( \text{pH} = 14 - 6.99 = 7.01 \). Wait, correction: Wait, \( K_a = 1.4 \times 10^{-4} \), so \( K_b = 1e-14 / 1.4e-4 ≈ 7.14e-11 \). Wait, moles of NaOH: \( 20.0 \, \text{mL} = 0.0200 \, \text{L} \), \( 4.0e-4 \, \text{mol/L} \) → \( 0.0200 * 4.0e-4 = 8.0e-6 \, \text{mol} \). Moles of lactic acid: same, so \( [\text{Lac}^-] = 8.0e-6 / 0.0550 ≈ 1.45e-4 \, \text{mol/L} \). Then \( K_b = x^2 / 1.45e-4 \) → \( x = \sqrt{7.14e-11 * 1.45e-4} = \sqrt{1.035e-14} ≈ 1.017e-7 \). Then \( \text{pOH} = 6.99 \), \( \text{pH} = 7.01 \). But wait, lactic acid is a weak acid, so at equivalence, the conjugate base is formed. Wait, maybe miscalculation. Wait, \( K_a = 1.4e-4 \), so \( K_b = 1e-14 / 1.4e-4 ≈ 7.14e-11 \). Then \( x = \sqrt{K_b * C} = \sqrt{7.14e-11 * 1.45e-4} = \sqrt{1.035e-14} ≈ 1.017e-7 \). So \( \text{OH}^- = 1.017e-7 \), \( \text{pOH} ≈ 6.99 \), \( \text{pH} ≈ 7.01 \). But wait, maybe I made a mistake in moles. Wait, lactic acid is monoprotic, so 1:1 with NaOH. So moles of lactic acid: \( 35.0 \, \text{mL} \) of \( C \), moles = \( 0.0350 * C \). Moles of NaOH: \( 0.0200 * 4.0e-4 = 8.0e-6 \). So \( 0.0350 * C = 8.0e-6 \) → \( C = 8.0e-6 / 0.0350 ≈ 2.286e-4 \, \text{mol/L} \). Then at equivalence, \( [\text{Lac}^-] = 8.0e-6 / (0.0350 + 0.0200) = 8.0e-6 / 0.0550 ≈ 1.45e-4 \, \text{mol/L} \). Then \( K_b = x^2 / 1.45e-4 \) → \( x = \sqrt{7.14e-11 * 1.45e-4} = 1.017e-7 \). So \( \text{pOH} = 6.99 \), \( \text{pH} = 7.01 \). Wait, but maybe the assumption \( x \ll 1.45e-4 \) is valid? Yes, since \( x = 1e-7 \) vs \( 1.45e-4 \). So pH ≈ 7.01. But wait, maybe I messed up \( K_a \). Wait, the problem says \( K_a = 1.4e-4 \), so lactic acid is a weak acid, conjugate base is weak base. So the pH at equivalence should be slightly basic? Wait, no: if \( K_a = 1.4e-4 \), \( pK_a = 3.85 \), so \( pK_b = 14 - 3.85 = 10.15 \), \( K_b = 10^{-10.15} ≈ 7.08e-11 \), which is close to our earlier calculation. Then \( [\text{OH}^-] = \sqrt{K_b * C} = \sqrt{7.08e-11 * 1.45e-4} ≈ \sqrt{1.027e-14} ≈ 1.013e-7 \), so \( \text{pOH} ≈ 6.99 \), \( \text{pH} ≈ 7.01 \). So the pH is approximately 7.01.

Answer:

\( \approx 2.10 \)

Problem 2: