What is the hydronium ion concentration of 1g of acetic acid in 100 ml of water?
Solution:
acetic acid = CH3COOH
hydronium ion = H3O+
The molar mass of CH3COOH is 60.052 g/mol
Therefore,
(1 g CH3COOH) × (1 mol CH3COOH / 60.052 g CH3COOH) = 0.01665 mol CH3COOH
Liters of CH3COOH solution = (100 mL soln) × (1 L / 1000 mL) = 0.1 L
Molarity of CH3COOH = Moles of CH3COOH / Liters of CH3COOH solution
Therefore,
Molarity of CH3COOH = (0.01665 mol) / (0.1 L) = 0.1665 mol/L = 0.1665 M
Acetic acid (CH3COOH) is a weak acid because it partially dissociates in an aqueous solution or it does not ionize completely in solution to produce hydronium ions (H3O+).
Equation representing the dissociation of CH3COOH:
CH3COOH(aq) + H2O(l) ⇋ H3O+(aq) + CH3COO−(aq)
ICE Table:
According to the ICE Table:
[H3O+] = [CH3COO−] = X
[CH3COOH] = 0.1665 − X ≈ 0.1665 M (assume that 0.1665 >> X)
The Ka expression for the dissociation of CH3COOH:
Ka = 1.75×10−5 (for CH3COOH)
Therefore,
1.75×10−5 = [X] × [X] / [0.1665]
2.914×10−6 = X2
X = (2.914×10−6)1/2 = 0.0017
[H3O+] = X = 0.0017 M = 1.7×10−3 M
[H3O+] = 1.7×10−3 M
Answer: The hydronium ion (H3O+) concentration is 1.7×10−3 M
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