Answer:
1. pH = 2,82
2. 3,20mL of 1,135M NaOH
3. pH = 3,25
Explanation:
The buffer of acetic acid (HC₂H₃O₂) is:
HC₂H₃O₂ ⇄ H⁺ + C₂H₃O₂⁻
The reaction of HC₂H₃O₂ with NaOH produce:
HC₂H₃O₂ + NaOH → C₂H₃O₂⁻ + Na⁺ + H₂O
And ka is defined as:
ka = [H⁺] [C₂H₃O₂⁻] / [HC₂H₃O₂] = 1,8x10⁻⁵ <em>(1)</em>
1. When in the solution you have just 0,13M HC₂H₃O₂ the concentrations in equilibrium will be:
[H⁺] = x
[C₂H₃O₂⁻] = x
[HC₂H₃O₂] = 0,13 - x
Replacing in (1)
[x] [x] / [0,13-x] = 1,8x10⁻⁵
x² = 2,34x10⁻⁶ - 1,8x10⁻⁵x
x² - 2,34x10⁻⁶ + 1,8x10⁻⁵x = 0
Solving for x:
x = - 0,0015 <em>(Wrong answer, there is no negative concentrations)</em>
x = 0,0015
As [H⁺] = x = 0,0015 and pH is -log [H⁺], pH of the solution is <em>2,82</em>
2. The equivalence point is reached when moles of HC₂H₃O₂ are equal to moles of NaOH. Moles of HC₂H₃O₂ are:
0,0466L × (0,078mol / L) = 3,63x10⁻³ moles of HC₂H₃O₂
In a 1,135M NaOH, these moles are reached with the addition of:
3,63x10⁻³ moles × (L / 1,135mol) = 3,20x10⁻³L = <em>3,20mL of 1,135M NaOH</em>
3. The initial moles of HC₂H₃O₂ are:
0,0172L × (0,128mol / L) = 2,20x10⁻³ moles of HC₂H₃O₂
As the addition of NaOH spent HC₂H₃O₂ producing C₂H₃O₂⁻. Moles of C₂H₃O₂⁻ are equal to moles of NaOH and moles of HC₂H₃O₂ are initial moles - moles of NaOH. That means:
0,46x10⁻³L NaOH × (0,155mol / L) = 7,13x10⁻⁵ moles of NaOH ≡ moles of C₂H₃O₂⁻
Final moles of HC₂H₃O₂ are:
2,20x10⁻³ - 7,13x10⁻⁵ = <em>2,2187x10⁻³ moles of HC₂H₃O₂</em>
Using Henderson-Hasselbalch formula:
pH = pka + log₁₀ [C₂H₃O₂⁻] / [HC₂H₃O₂]
Where pka is -log ka = 4,74. Replacing:
pH = 4,74 + log₁₀ [7,13x10⁻⁵] / [2,2187x10⁻³ ]
<em>pH = 3,25</em>
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I hope it helps!
