What is the ph of solution when 50 mL of 1 M CH3COOH (acetic acid) is dissolve in 8.80 g, 12.0 g, 15.5 g, CH3COONA (sodium acetate)?
Answers
Answer:
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When 50 ml of 1 M acetic acid is added to 8.80 g, 12 g, and 15.5 g of CH3COONa, the pH of the solution is 6.37, 6.5, and 6.61 respectively.
• Given,
Molarity (M) of acetic acid = 1 M = 1 mole / L
=> Molarity in 1000 ml of the solution = 1
Volume of acetic acid = 50 ml
=> Molarity of acetic acid in 50 ml of the solution = (1/1000) × 50 = 0.05 M
• The pH of a solution of salt and acid can be determined by the Henderson-Hasselbach equation, which goes as follows :
pH = pKa + log ( [Salt] / [Acid] )
where pKa is the strength of acid,
[Salt] is the concentration of salt,
[Acid] is the concentration of acid.
• pKa of acetic acid is 4.74 .
Here, the salt is CH3COONa, and the acid is acetic acid.
• Case 1 :
Amount of CH3COONa added = 8.80 g
=> Molecular weight of CH3COONa = 82 g
=> Molarity of CH3COONa in 500 ml of solution = number of moles / Volume in L
=> Molarity = (8.80/82) moles/0.05 L
=> Molarity = 0.107 / 0.05 M
=> Molarity = 2.14 M
Therefore, applying Henderson-Hasselbach equation, we get,
pH = 4.74 + log ( [2.14] / [0.05] )
=> pH = 4.74 + log 42.8
=> PH = 4.74 + 1.63
=> pH = 6.37
• Case 2 :
Amount of CH3COONa added = 12 g
=> Molecular weight of CH3COONa = 82 g
=> Molarity of CH3COONa in 500 ml of solution = number of moles / Volume in L
=> Molarity = (12/82) moles/0.05 L
=> Molarity = 0.146 / 0.05 M
=> Molarity = 2.92 M
Therefore, applying Henderson-Hasselbach equation, we get,
pH = 4.74 + log ( [2.92] / [0.05] )
=> pH = 4.74 + log 58.4
=> PH = 4.74 + 1.76
=> pH = 6.5
• Case 3 :
Amount of CH3COONa added = 15.5 g
=> Molecular weight of CH3COONa = 82 g
=> Molarity of CH3COONa in 500 ml of solution = number of moles / Volume in L
=> Molarity = (15.5/82) moles/0.05 L
=> Molarity = 0.189 / 0.05 M
=> Molarity = 3.78 M
Therefore, applying Henderson-Hasselbach equation, we get,
pH = 4.74 + log ( [3.78] / [0.05] )
=> pH = 4.74 + log 75.6
=> PH = 4.74 + 1.87
=> pH = 6.61