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Misha Larkins [42]
3 years ago
13

Starting with 0.250L of a buffer solution containing 0.250 M benzoic acid (C 6H 5COOH) and 0.20 M sodium benzoate (C 6H 5COONa),

what will the pH of the solution be after the addition of 25.0 mL of 0.100M HCl? (K a (C 6H 5COOH) = 6.5 x 10 -5)
Chemistry
1 answer:
Zigmanuir [339]3 years ago
5 0

Answer:

pH = 4.05

Explanation:

The pH of the benzoic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

<em>Where pKa is -logKa = 4.187</em>

pH = 4.187 + log [Sodium Benzoate] / [Benzoic Acid]

<em>Where [] can be understood as moles of each specie.</em>

Thus, to find pH of the buffer we need to calculate moles of benzoic acid and sodium benzoate.

<em>Initial moles:</em>

<em />

Initial moles of benzoic acid and sodium benzoate are:

Acid: 250mL = 0.250L ₓ (0.250 moles / L) = <em>0.0625 moles of benzoic acid</em>

Benzoate : 250mL = 0.20L ₓ (0.250 moles / L) = <em>0.050 moles of sodium benzoate</em>

<em>Moles after reaction:</em>

Now, 0.0250L×(0.100mol/L) = 0.0025 moles of HCl are added to the buffer reacting with sodium benzoate, C₆H₅COONa, producing more benzoic acid, as follows:

HCl + C₆H₅COONa → C₆H₅COOH + NaCl

That means after reaction moles of both species are:

Benzoic acid: 0.0625 mol + 0.0025mol (Moles produced) = 0.065 moles

Sodium Benzoate: 0.050mol - 0.0025mol (Moles that react) = 0.0475 moles

Replacing in H-H equation:

pH = 4.187 + log [0.0475] / [0.065]

pH = 4.05

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7s^2 5f^4

This notation can be written in core notation or noble gas notation by replacing the

1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^10 4p^6 5s^2 4d^10 5p^6 6s^2 4e^14 5d^10 6p^6

7s^2 5f^4

with the noble gas [Rn].

[Rn]7s25f4

N[nitrogen] The full electron configuration for nitrogen is 1s^2 2s^2 2p^3.

Ti[titanium] Ti2+:[Ar]3d^2

Ti:1s^2 2s^2 2p^6 3s^2 3p^6 3d^2 4s^2

1s^2 2s^2 2p^6 3s^2 3p^5 = 17 electrons

(1) electron gain will result to a

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Hg[mercury] You should then find its atomic number is 80. It has a Xe core, so in shorthand notation, you can include [Xe]instead of

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for 54 electrons. For the 6th row of the periodic table, we introduce the 4f orbitals, and proceed to atoms having occupied 5d orbitals. We, as usual, have the ns orbitals, and n=? for the 6th period?

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Explanation:

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6 0
2 years ago
How many moles in 2.21 x10E24 atoms of aluminum
lisov135 [29]
<h3>Answer:</h3>

3.67 mol Al

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

2.21 × 10²⁴ atoms Al

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

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 2.21 \cdot 10^{24} \ atoms \ Al(\frac{1 \ mol \ Al}{6.022 \cdot 10^{23} \ atoms \ Al})
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<u>Step 4: Check</u>

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

3.66988 mol Al ≈ 3.67 mol Al

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Vinvika [58]

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