What is the pH of a solution prepared by adding 4.000 grams of NaOH to a 50.00 mL of a buffer that is 2.00 molar both in acetic acid and sodium acetate? (Assume final volume is still 50.00 mL)
* chem - DrBob222, Wednesday, April 21, 2010 at 10:52pm
* chem - phil, Wednesday, April 21, 2010 at 10:59pm
i did .1 mols of AC- + .1 mols of OH gives me .2 mols of AC.
the .2 mols of AC reacts with water
AC- + H20---> OH- + HAC
so the M is .2/.05 L = 4M
to solve for the OH conc i did X^2 / 4 = 5.6 x 10^-10. and solving for the pH i get 9.675
is this right?
Yes, it's right. I answered below at the original post.
To find the pH of the solution, you need to follow these steps:
Step 1: Calculate the number of moles of NaOH added.
Given:
Mass of NaOH = 4.000 grams
To find the number of moles, you can use the formula:
moles = mass / molar mass
The molar mass of NaOH is:
Sodium (Na) = 22.99 g/mol
Oxygen (O) = 16.00 g/mol
Hydrogen (H) = 1.01 g/mol
Molar mass of NaOH = Na (22.99) + O (16.00) + H (1.01) = 39.00 g/mol
moles of NaOH = 4.000 grams / 39.00 g/mol = 0.1026 moles
Step 2: Calculate the total moles of acetic acid and sodium acetate in the buffer solution.
Given:
Volume of the buffer solution = 50.00 mL
Molarity of the buffer solution = 2.00 M (= 2.00 moles/L)
To calculate the moles, you can use the formula:
moles = molarity × volume (in liters)
moles of acetic acid and sodium acetate = 2.00 M × (50.00 mL / 1000 mL/ L) = 0.1000 moles
Step 3: Calculate the total moles of acetic acid after the reaction with NaOH.
Since NaOH is a strong base, it will react with acetic acid to form water and acetate ions (AC-).
The balanced equation for the reaction between NaOH and acetic acid is:
CH3COOH + NaOH → H2O + CH3COONa
From the equation, you can see that 1 mole of acetic acid reacts with 1 mole of NaOH to form 1 mole of acetate ions.
moles of acetate ions (AC-) = moles of NaOH added = 0.1026 moles
Step 4: Calculate the molarity of the acetate ions.
Given:
Volume of the solution = 50.00 mL
Total moles of acetate ions = 0.1026 moles
To calculate the molarity, you can use the formula:
Molarity = moles / volume (in liters)
Molarity of acetate ions = 0.1026 moles / (50.00 mL / 1000 mL/ L) = 2.052 M
Step 5: Calculate the concentration of hydroxide ions (OH-) produced by the reaction between acetate ions and water.
Given:
Molarity of acetate ions = 2.052 M
Since the reaction between acetate ions and water produces equal amounts of hydroxide ions (OH-) and acetic acid (HAC), the molarity of hydroxide ions would be the same as the molarity of acetate ions.
Molarity of hydroxide ions = 2.052 M
Step 6: Calculate the pOH of the solution.
pOH = -log10 (OH- concentration)
pOH = -log10 (2.052) = 0.687
Step 7: Calculate the pH of the solution.
pH + pOH = 14 (at 25 degrees Celsius)
pH = 14 - pOH
pH = 14 - 0.687
pH = 13.313
Therefore, the pH of the solution is approximately 13.313.