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Vilka [71]
2 years ago
6

Jacob carelessly added only 40.0 mL (instead of the recommended 50.0 mL) of 1.1 M HCl to the 50.0 mL of 1.0 M NaOH. Explain the

consequence of the error.
Chemistry
1 answer:
Sati [7]2 years ago
8 0

Answer:

Explanation:

mole of NaOH present = molarity x volume

                                     = 1.0 X 0.05 = 0.05 mole

<em>Recommended mole of HCl </em>= 1.1 x 0.05 = 0.055

<em>Mole of HCl carelessly added by Jacob </em>= 1.1 x 0.04 = 0.044

From the equation of reaction:

HCl + NaOH ----> NaCl + H2O

The ratio of mole of HCl to that of NaOH for a complete neutralization reaction is 1:1. However, the recommended mole of HCl (0.055 mole) is more than the mole of NaOH (0.05 mole). <u>Hence, the recommended endpoint of the reaction is supposed to be acidic.</u>

The mole of HCl added by Jacob (0.044) is short of the recommended amount (0.055) and also short of the amount required for a neutral endpoint (0.05). <u>This means that the endpoint will have an excess amount of NaOH and as such, basic instead of the desired acidic endpoint.</u>

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Answer:

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General Formulas and Concepts:

<u>Atomic Structure</u>

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

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 3.50 × 10²⁴ molecules Cl₂

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<u>Step 2: Identify Conversions</u>

Avogadro's Number

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Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 3.50 \cdot 10^{24} \ molecules \ Cl_2(\frac{1 \ mol \ Cl_2}{6.022 \cdot 10^{23} \ molecules \ Cl_2})(\frac{70.9 \ g \ Cl_2}{1 \ mol \ Cl_2})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 412.072 \ g \ Cl_2

<u>Step 4: Check</u>

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

412.072 g Cl₂ ≈ 412 g Cl₂

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