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Mariana [72]
3 years ago
10

If 20.0 ml of glacial acetic acid (pure hc2h3o2) is diluted to 1.40 l with water, what is the ph of the resulting solution? the

density of glacial acetic acid is 1.05 g/ml.
Chemistry
1 answer:
Y_Kistochka [10]3 years ago
8 0

By using the formula, mass = density x volume, we calculate mass in grams

 20.0 mL CH₃COOH x (1.05 g / mL) = 21.0 g CH₃COOH 

To find the moles, molar mass of CH₃COOH = 60.05g/mol<span>

21.0 g </span>CH₃COOH x (1 mole CH₃COOH / 60.05 g CH₃COOH) = 0.350 moles CH₃COOH 

To find molarity,<span>

[</span>CH₃COOH] = moles CH₃COOH / L of solution = 0.350 / 1.40 = 0.250 M CH₃COOH<span> 

When </span>CH₃COOH is dissolved in water, it produces small and equal amounts of H₃O⁺+ and C₂H₃O₂⁻. 

<span>
Molarity ,         </span>CH₃COOH<span> + H</span>₂O <==> H₃O⁺ + C₂H₃O₂⁻ 

<span> <span>Initial                      0.250                          0           0 </span>
Change                      -x                            x            x 
Equilibrium            0.250-x                        x            x 

K</span>ₐ = [H₃O⁺][C₂H₃O₂⁻] / [HC₂H₃O₂] = (x)(x) / (0.250-x) = 1.8 x 10⁻⁵

<span>Since K</span>ₐ is relatively small, we can neglect the -x term after 0.250 to simplify 

<span>x</span>² / 0.250 = 1.8 x 10⁻⁵ 

x² = 4.5 x 10⁻⁶ 

<span> x = 2.1 x 10</span>⁻³<span> = [H</span>₃O⁺] 

pH = -log [H₃O⁺] = -log (2.1 x 10⁻³) = 2.68

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