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aleksklad [387]
3 years ago
15

What is the ph of a buffer prepared by adding 0.607 mol of the weak acid ha to 0.609 mol of naa in 2.00 l of solution? the disso

ciation constant ka of ha is 5.66×10−7?
Chemistry
1 answer:
Paraphin [41]3 years ago
6 0

Given:

0.607 mol of the weak acid

0.609 naa

2.00 liters of solution

 

The solution for finding the ph of a buffer:

[HA] = 0.607 / 2.00 = 0.3035 M 
[A-]= 0.609/ 2.00 = 0.3045 M 
pKa = 6.25 

pH = 6.25 + log 0.3045/ 0.3035 = 6.25 is the ph buffer prepared.

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1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this
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<h3>Answer:</h3>

0.6 g NaCl

<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>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

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2 years ago
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