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nasty-shy [4]
3 years ago
12

Calculate the [ H + ] and pH of a 0.0040 M butanoic acid solution. The K a of butanoic acid is 1.52 × 10 − 5 . Use the method of

successive approximations in your calculations.
Chemistry
1 answer:
stellarik [79]3 years ago
4 0

Answer:

Ka= [H+][But-]/H[Hbut]

Let k moles of H+ be formed

1.52*10^-5=k^2/0.004-k

K^2=0.004*1.52*10^-5-1.52*10^-5k

K^2- 1.52*10^-5k-6.08*10^-8=0

Solving this quadratic equation.

K=7.6*10^-6

PH=-log(7.6*10^-6)=5.2

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Answer:

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Explanation:

Please see the step-by-step solution in the picture attached below.

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s2008m [1.1K]

Answer:

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General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Reading a Periodic Table
  • Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

Explanation:

<u>Step 1: Define</u>

97.3 g CO₂

<u>Step 2: Define conversions</u>

Avogadro's Number

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Molar Mass of O - 16.00 g/mol

Molar Mass of CO₂ - 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

97.3 \ g \ CO_2(\frac{1 \ mol \ CO_2}{44.01 \ g \ CO_2} )(\frac{6.022 \cdot 10^{23} \ molecules \ CO_2}{1 \ mol \ CO_2} ) = 1.33138 × 10²⁴ molecules CO₂

<u>Step 4: Check</u>

<em>We are given 3 sig figs. Follow sig fig rules.</em>

1.33138 × 10²⁴ molecules CO₂ ≈ 1.33 × 10²⁴ molecules CO₂

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