Question

Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH. What is the final pH? The Ka of acetic acid is 1.770 × 10-5

Answer 1

To calculate the final pH of the solution, you need to consider the reaction between acetic acid (CH3COOH) and NaOH. The concentration of OH- ions in the final solution determines the pH, which can be calculated using the formula pH = -log10(H+ concentration). Using the given information, we can calculate the concentration of OH- ions, and then convert that to the concentration of H+ ions to find the pH of the solution.

Step-by-step explanation:

To calculate the final pH of the solution, we need to consider the reaction between acetic acid (CH3COOH) and NaOH. The reaction between the two compounds forms sodium acetate (CH3COONa) and water.

The balanced equation for the reaction is:

CH3COOH + NaOH → CH3COONa + H2O

Since NaOH is a strong base, it completely dissociates in water to produce OH- ions. These OH- ions react with the acetic acid to form water. Therefore, the concentration of OH- ions will determine the pH of the final solution.

Using the given information, we can calculate the moles of acetic acid and OH- ions:

Moles of acetic acid = volume of acetic acid solution (L) x concentration of acetic acid (M)

= 0.200 L x 0.5000 M = 0.100 moles

Moles of OH- ions = volume of NaOH solution (L) x concentration of NaOH (M)

= 0.100 L x 0.5000 M = 0.050 moles

Since the mole ratio between acetic acid and OH- ions is 1:1, there will be an equal number of moles of OH- ions and acetic acid in the solution after the reaction is complete.

Therefore, the concentration of OH- ions in the final solution is:

Concentration of OH- ions = Moles of OH- ions / volume of solution (L)

= 0.050 moles / 0.300 L = 0.167 M

To calculate the final pH, we can use the formula: pH = -log10(H+ concentration)

Since in water, OH- ions and H+ ions are inversely proportional, we can calculate the concentration of H+ ions using:

H+ concentration = Kw / OH- concentration

Where Kw is the ion product of water and is equal to 1.0 x 10^-14 at 25°C.

Therefore, H+ concentration = 1.0 x 10^-14 / 0.167 M = 5.99 x 10^-14 M

Finally, calculating the pH:

pH = -log10(5.99 x 10^-14) ≈ 13.22

Answer 2

The final pH is 3.80

Step-by-step explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80