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What amount of solid sodium acetate be added into 1 litre of the 0.1 M CH3COOH solution so that the resulting solution has pH almost equal to pKa (CH3COOH) = 4.74
- a)1.224 g
- b)5.232 g
- c)10 g
- d)14.924 g
Correct answer is option 'D'. Can you explain this answer?
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To calculate the amount of solid sodium acetate needed to achieve a pH almost equal to pKa (CH3COOH), we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the acetate ion (CH3COO-) and [HA] represents the concentration of acetic acid (CH3COOH).
We are given that the pKa of acetic acid is 4.74 and the concentration of acetic acid is 0.1 M. Since the initial solution only contains acetic acid, the initial concentration of acetate ion is 0 M.
Let's assume that x grams of solid sodium acetate (CH3COONa) are added to the solution. Sodium acetate dissociates completely in water, so the concentration of acetate ion after adding the solid sodium acetate will be x mol/L.
- Initial concentration of acetic acid ([HA]): 0.1 M
- Initial concentration of acetate ion ([A-]): 0 M
- Final concentration of acetic acid ([HA]): 0.1 M
- Final concentration of acetate ion ([A-]): x M
Substituting these values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(x/0.1)
To achieve a pH almost equal to the pKa, we want the logarithmic term to be zero. This means:
log(x/0.1) = 0
Solving for x:
x/0.1 = 1
x = 0.1
Therefore, we need to add 0.1 mol/L of sodium acetate to the solution.
To convert this to grams, we need to calculate the molar mass of sodium acetate:
- Sodium (Na): 22.99 g/mol
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
Total molar mass of sodium acetate (CH3COONa) = 82.03 g/mol
Now we can calculate the mass of solid sodium acetate needed:
mass = moles × molar mass
mass = 0.1 mol/L × 82.03 g/mol
mass ≈ 8.203 g
Therefore, the correct answer is option 'D' - 14.924 g.
pH = pKa + log([A-]/[HA])
In this equation, [A-] represents the concentration of the acetate ion (CH3COO-) and [HA] represents the concentration of acetic acid (CH3COOH).
We are given that the pKa of acetic acid is 4.74 and the concentration of acetic acid is 0.1 M. Since the initial solution only contains acetic acid, the initial concentration of acetate ion is 0 M.
Let's assume that x grams of solid sodium acetate (CH3COONa) are added to the solution. Sodium acetate dissociates completely in water, so the concentration of acetate ion after adding the solid sodium acetate will be x mol/L.
- Initial concentration of acetic acid ([HA]): 0.1 M
- Initial concentration of acetate ion ([A-]): 0 M
- Final concentration of acetic acid ([HA]): 0.1 M
- Final concentration of acetate ion ([A-]): x M
Substituting these values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(x/0.1)
To achieve a pH almost equal to the pKa, we want the logarithmic term to be zero. This means:
log(x/0.1) = 0
Solving for x:
x/0.1 = 1
x = 0.1
Therefore, we need to add 0.1 mol/L of sodium acetate to the solution.
To convert this to grams, we need to calculate the molar mass of sodium acetate:
- Sodium (Na): 22.99 g/mol
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.01 g/mol
- Oxygen (O): 16.00 g/mol
Total molar mass of sodium acetate (CH3COONa) = 82.03 g/mol
Now we can calculate the mass of solid sodium acetate needed:
mass = moles × molar mass
mass = 0.1 mol/L × 82.03 g/mol
mass ≈ 8.203 g
Therefore, the correct answer is option 'D' - 14.924 g.
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What amount of solid sodium acetate be added into 1 litre of the 0.1 M CH3COOH solution so that the resulting solution has pH almost equal to pKa (CH3COOH) = 4.74a)1.224gb)5.232 gc)10 gd)14.924gCorrect answer is option 'D'. Can you explain this answer?
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