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Lisa [10]
2 years ago
7

a solution of household vinegar is to be analyzed. A pipet is used to measure out 10 ml of the vinegar which is placed in a 250m

l volumetric flask. Distilled water is added until the total volume of solution is 250 ml portion of the diluted solution is measured out with a pipet and titrated with a standard solution of sodium hydroxide. The neutralization reaction is as follows: HC2H3O2 (aq) + OH- (aq) --> C2H3O2- (aq) + H2O (l)It is found that 16.7 mL of 0.0500 M NaOH is needed to titrate 25.0 mL of the diluted vinegar. Calculate the molarity of the diluted vinegar.
Chemistry
1 answer:
DerKrebs [107]2 years ago
6 0

Answer:

0.0344 M

Explanation:

  • HC₂H₃O₂ (aq) + OH⁻ (aq) --> C₂H₃O₂⁻ (aq) + H₂O (l)

Because one mol of vinegar (acetic acid) reacts with one mol of NaOH, we can use the formula

  • C₁V₁=C₂V₂

Where C₁ and V₁ refer to the concentration and volume of vinegar, and C₂ and V₂ to those of NaOH. We're given V₁, C₂ and V₂; so we <em>solve for C₁</em>:

  • C₁ = C₂V₂ / V₁
  • C₁ = 0.0500 M * 16.7 mL / 25.0 mL
  • C₁ = 0.0344 M
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Answer:

[H₃O⁺] = 4.3 × 10⁻¹² mol·L⁻¹; [OH⁻] = 2.4 × 10⁻³ mol·L⁻¹;

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Explanation:

The chemical equation is

\rm NH$_{3}$ + \text{H}$_{2}$O \, \rightleftharpoons \,$ NH$_{4}^{+}$ + \,\text{OH}$^{-}$

For simplicity, let's re-write this as

\rm B + H$_{2}$O \, \rightleftharpoons\,$ BH$^{+}$ + OH$^{-}$

1. Calculate [OH]⁻

(a) Set up an ICE table.  

   B + H₂O ⇌ BH⁺ + OH⁻

0.310               0        0

   -x                  +x      +x

0.310-x               x        x

K_{\text{b}} = \dfrac{\text{[BH}^{+}]\text{[OH}^{-}]}{\text{[B]}} = 1.8 \times 10^{-5}\\\\\dfrac{x^{2}}{0.100 - x} = 1.8 \times 10^{-5}

Check for negligibility:

\dfrac{0.310}{1.8 \times 10^{-5}} = 17 000 > 400\\\\x \ll 0.310

(b) Solve for [OH⁻]

\dfrac{x^{2}}{0.310} = 1.8 \times 10^{-5}\\\\x^{2} = 0.310 \times 1.8 \times 10^{-5}\\x^{2} = 5.58 \times 10^{-6}\\x = \sqrt{5.58 \times 10^{-6}}\\x = \text{[OH]}^{-} = \mathbf{2.4 \times 10^{-3}} \textbf{ mol/L}

2. Calculate the pOH

\text{pOH} = -\log \text{[OH}^{-}] = -\log(2.4 \times 10^{-3}) = \mathbf{2.6}

3. Calculate the pH

\text{pH} = 14.00 - \text{pOH} = 14.00 - 2.6 = \mathbf{11.4}

4 Calculate [H₃O⁺]

\text{H$_{3}$O$^{+}$} = 10^{-\text{pH}} = 10^{-11.4} = \mathbf{4.2 \times 10^{-12}} \textbf{ mol/L}

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Answer:

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Answer:

4.16L

Explanation:

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The rule of thumb for acidity with relation to pKa values is that:

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In aniline , the stabilized benzene ring attached to NH₂ makes it a less electron withdrawing group compared to the straight chains structure found in secondary amine where electron are easily withdraw by nucleophilic substitution reactions.

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