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20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)?
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?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The p...
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?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The p...
Calculating the pH of the Solution
To determine the pH of the solution after mixing 20 ml of 0.2 M NaOH with 50 ml of 0.2 M acetic acid, we need to consider the reaction between the two compounds. NaOH is a strong base, while acetic acid is a weak acid. The reaction between a strong base and a weak acid is known as a neutralization reaction.
Neutralization Reaction
NaOH + CH3COOH -> CH3COONa + H2O
In this reaction, NaOH reacts with acetic acid (CH3COOH) to form sodium acetate (CH3COONa) and water (H2O).
Calculating the Moles of NaOH and Acetic Acid
To calculate the pH of the solution, we need to determine the moles of NaOH and acetic acid used.
Moles of NaOH:
Moles = Concentration (M) * Volume (L)
Moles of NaOH = 0.2 M * 0.02 L
Moles of NaOH = 0.004 moles
Moles of Acetic Acid:
Moles = Concentration (M) * Volume (L)
Moles of acetic acid = 0.2 M * 0.05 L
Moles of acetic acid = 0.01 moles
Determining the Limiting Reactant
To determine the limiting reactant, we compare the moles of NaOH and acetic acid used. The reactant that is completely consumed is the limiting reactant.
In this case, the moles of NaOH (0.004 moles) are lesser than the moles of acetic acid (0.01 moles). Therefore, NaOH is the limiting reactant.
Calculating the Moles of Sodium Acetate
Since NaOH is the limiting reactant, all of it will be converted to sodium acetate.
Moles of sodium acetate = Moles of NaOH = 0.004 moles
Calculating the Concentration of Sodium Acetate
Sodium acetate is formed from the reaction between NaOH and acetic acid. The total volume of the solution after mixing is 20 ml + 50 ml = 70 ml.
Concentration (M) = Moles / Volume (L)
Concentration of sodium acetate = 0.004 moles / 0.07 L
Concentration of sodium acetate = 0.057 M
Calculating the pH of Sodium Acetate Solution
To calculate the pH of the sodium acetate solution, we need to consider the dissociation of sodium acetate in water. Sodium acetate is a salt that dissociates into its ions.
CH3COONa -> CH3COO- + Na+
The dissociation of sodium acetate does not produce any H+ ions, which means it does not contribute to the acidity of the solution. Therefore, the pH of the sodium acetate solution is determined by the hydrolysis of the acetate ion (CH3COO-).
Hydrolysis of Acetate Ion
CH3COO- + H2O -> CH3COOH +
To determine the pH of the solution after mixing 20 ml of 0.2 M NaOH with 50 ml of 0.2 M acetic acid, we need to consider the reaction between the two compounds. NaOH is a strong base, while acetic acid is a weak acid. The reaction between a strong base and a weak acid is known as a neutralization reaction.
Neutralization Reaction
NaOH + CH3COOH -> CH3COONa + H2O
In this reaction, NaOH reacts with acetic acid (CH3COOH) to form sodium acetate (CH3COONa) and water (H2O).
Calculating the Moles of NaOH and Acetic Acid
To calculate the pH of the solution, we need to determine the moles of NaOH and acetic acid used.
Moles of NaOH:
Moles = Concentration (M) * Volume (L)
Moles of NaOH = 0.2 M * 0.02 L
Moles of NaOH = 0.004 moles
Moles of Acetic Acid:
Moles = Concentration (M) * Volume (L)
Moles of acetic acid = 0.2 M * 0.05 L
Moles of acetic acid = 0.01 moles
Determining the Limiting Reactant
To determine the limiting reactant, we compare the moles of NaOH and acetic acid used. The reactant that is completely consumed is the limiting reactant.
In this case, the moles of NaOH (0.004 moles) are lesser than the moles of acetic acid (0.01 moles). Therefore, NaOH is the limiting reactant.
Calculating the Moles of Sodium Acetate
Since NaOH is the limiting reactant, all of it will be converted to sodium acetate.
Moles of sodium acetate = Moles of NaOH = 0.004 moles
Calculating the Concentration of Sodium Acetate
Sodium acetate is formed from the reaction between NaOH and acetic acid. The total volume of the solution after mixing is 20 ml + 50 ml = 70 ml.
Concentration (M) = Moles / Volume (L)
Concentration of sodium acetate = 0.004 moles / 0.07 L
Concentration of sodium acetate = 0.057 M
Calculating the pH of Sodium Acetate Solution
To calculate the pH of the sodium acetate solution, we need to consider the dissociation of sodium acetate in water. Sodium acetate is a salt that dissociates into its ions.
CH3COONa -> CH3COO- + Na+
The dissociation of sodium acetate does not produce any H+ ions, which means it does not contribute to the acidity of the solution. Therefore, the pH of the sodium acetate solution is determined by the hydrolysis of the acetate ion (CH3COO-).
Hydrolysis of Acetate Ion
CH3COO- + H2O -> CH3COOH +
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?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)?
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?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)?.
?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about ?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for ?20 ml of 0.2 M NaOh is added to 50 ml of 0.2 M of acetic acid . The pH of this solution after mixing is (ka =1.8×10^-5)?.
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